Ответы к странице 192
788. Решите уравнение:
1) $\frac{60}{x} - \frac{60}{x + 10} = \frac{1}{5}$;
2) $\frac{x}{x + 2} + \frac{x + 2}{x - 2} = \frac{16}{x^2 - 4}$;
3) $\frac{9}{x + 3} + \frac{14}{x - 3} = \frac{24}{x}$;
4) $\frac{2y + 3}{2y + 2} - \frac{y + 1}{2y - 2} + \frac{1}{y^2 - 1} = 0$;
5) $\frac{3x}{x^2 - 10x + 25} - \frac{x - 3}{x^2 - 5x} = \frac{1}{x}$;
6) $\frac{x - 20}{x^2 + 10x} + \frac{10}{x^2 - 100} - \frac{5}{x^2 - 10x} = 0$.
Решение:
1) $\frac{60}{x} - \frac{60}{x + 10} = \frac{1}{5}$
x ≠ 0
и
x + 10 ≠ 0
x ≠ −10
$\frac{60}{x} - \frac{60}{x + 10} - \frac{1}{5} = 0$ | * 5x(x + 10)
300(x + 10) − 300x − x(x + 10) = 0
$300x + 3000 - 300x - x^2 - 10x = 0$
$-x^2 - 10x + 3000 = 0$ | * (−1)
$x^2 + 10x - 3000 = 0$
$D = b^2 - 4ac = 10^2 - 4 * 1 * (-3000) = 100 + 12000 = 12100 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-10 + \sqrt{12100}}{2 * 1} = \frac{-10 + 110}{2} = \frac{100}{2} = 50$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-10 - \sqrt{12100}}{2 * 1} = \frac{-10 - 110}{2} = \frac{-120}{2} = -60$
Ответ: −60 и 50
2) $\frac{x}{x + 2} + \frac{x + 2}{x - 2} = \frac{16}{x^2 - 4}$
$\frac{x}{x + 2} + \frac{x + 2}{x - 2} = \frac{16}{(x - 2)(x + 2)}$
x + 2 ≠ 0
x ≠ −2
и
x − 2 ≠ 0
x ≠ 2
$\frac{x}{x + 2} + \frac{x + 2}{x - 2} - \frac{16}{(x - 2)(x + 2)} = 0$ | * (x − 2)(x + 2)
$x(x - 2) + (x + 2)^2 - 16 = 0$
$x^2 - 2x + x^2 + 4x + 4 - 16 = 0$
$2x^2 + 2x - 12 = 0$ | : 2
$x^2 + x - 6 = 0$
$D = b^2 - 4ac = 1^2 - 4 * 1 * (-6) = 1 + 24 = 25 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-1 + \sqrt{25}}{2 * 1} = \frac{-1 + 5}{2} = \frac{4}{2} = 2$ − не удовлетворяет условию, так как x ≠ 2.
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-1 - \sqrt{25}}{2 * 1} = \frac{-1 - 5}{2} = \frac{-6}{2} = -3$
Ответ: −3
3) $\frac{9}{x + 3} + \frac{14}{x - 3} = \frac{24}{x}$
x + 3 ≠ 0
x ≠ −3
и
x − 3 ≠ 0
x ≠ 3
и
x ≠ 0
$\frac{9}{x + 3} + \frac{14}{x - 3} - \frac{24}{x} = 0$ | * x(x + 3)(x − 3)
$9x(x - 3) + 14x(x + 3) - 24(x^2 - 9) = 0$
$9x^2 - 27x + 14x^2 + 42x - 24x^2 + 216 = 0$
$-x^2 + 15x + 216 = 0$ | * (−1)
$x^2 - 15x - 216 = 0$
$D = b^2 - 4ac = (-15)^2 - 4 * 1 * (-216) = 225 + 864 = 1089 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{15 + \sqrt{1089}}{2 * 1} = \frac{15 + 33}{2} = \frac{48}{2} = 24$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{15 - \sqrt{1089}}{2 * 1} = \frac{15 - 33}{2} = \frac{-18}{2} = -9$
Ответ: −9 и 24
4) $\frac{2y + 3}{2y + 2} - \frac{y + 1}{2y - 2} + \frac{1}{y^2 - 1} = 0$
$\frac{2y + 3}{2(y + 1)} - \frac{y + 1}{2(y - 1)} + \frac{1}{(y - 1)(y + 1)} = 0$
y + 1 ≠ 0
y ≠ −1
и
y − 1 ≠ 0
y ≠ 1
$\frac{2y + 3}{2(y + 1)} - \frac{y + 1}{2(y - 1)} + \frac{1}{(y - 1)(y + 1)} = 0$ | * 2(y − 1)(y + 1)
$(2y + 3)(y - 1) - (y + 1)^2 + 2 = 0$
$2y^2 + 3y - 2y - 3 - (y^2 + 2y + 1) + 2 = 0$
$2y^2 + y - 1 - y^2 - 2y - 1 = 0$
$y^2 - y - 2 = 0$
$D = b^2 - 4ac = (-1)^2 - 4 * 1 * (-2) = 1 + 8 = 9 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{1 + \sqrt{9}}{2 * 1} = \frac{1 + 3}{2} = \frac{4}{2} = 2$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{1 - \sqrt{9}}{2 * 1} = \frac{1 - 3}{2} = \frac{-2}{2} = -1$ − не удовлетворяет условию, так как y ≠ −1.
Ответ: 2
5) $\frac{3x}{x^2 - 10x + 25} - \frac{x - 3}{x^2 - 5x} = \frac{1}{x}$
$\frac{3x}{(x - 5)^2} - \frac{x - 3}{x(x - 5)} = \frac{1}{x}$
$(x - 5)^2 ≠ 0$
x − 5 ≠ 0
x ≠ 5
и
x ≠ 0
$\frac{3x}{(x - 5)^2} - \frac{x - 3}{x(x - 5)} - \frac{1}{x} = 0$ | * $x(x - 5)^2$
$3x^2 - (x - 3)(x - 5) - (x - 5)^2 = 0$
$3x^2 - (x^2 - 3x - 5x + 15) - (x^2 - 10x + 25) = 0$
$3x^2 - x^2 + 8x - 15 - x^2 + 10x - 25 = 0$
$x^2 + 18x - 40 = 0$
$D = b^2 - 4ac = 18^2 - 4 * 1 * (-40) = 324 + 160 = 484 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-18 + \sqrt{484}}{2 * 1} = \frac{-18 + 22}{2} = \frac{4}{2} = 2$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-18 - \sqrt{484}}{2 * 1} = \frac{-18 - 22}{2} = \frac{-40}{2} = -20$
Ответ: −20 и 2
6) $\frac{x - 20}{x^2 + 10x} + \frac{10}{x^2 - 100} - \frac{5}{x^2 - 10x} = 0$
$\frac{x - 20}{x(x + 10)} + \frac{10}{(x - 10)(x + 10)} - \frac{5}{x(x - 10)} = 0$
x − 10 ≠ 0
x ≠ 10
и
x + 10 ≠ 0
x ≠ −10
и
x ≠ 0
$\frac{x - 20}{x(x + 10)} + \frac{10}{(x - 10)(x + 10)} - \frac{5}{x(x - 10)} = 0$ | * x(x − 10)(x + 10)
(x − 20)(x − 10) + 10x − 5(x + 10) = 0
$x^2 - 20x - 10x + 200 + 10x - 5x - 50 = 0$
$x^2 - 25x + 150 = 0$
$D = b^2 - 4ac = (-25)^2 - 4 * 1 * 150 = 625 + 600 = 25 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{25 + \sqrt{25}}{2 * 1} = \frac{25 + 5}{2} = \frac{30}{2} = 15$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{25 - \sqrt{25}}{2 * 1} = \frac{25 - 5}{2} = \frac{20}{2} = 10$ − не удовлетворяет условию, так как x ≠ 10.
Ответ: 15
789. При каком значении переменной:
1) сумма дробей $\frac{24}{x - 2}$ и $\frac{16}{x + 2}$ равна 3;
2) значение дроби $\frac{42}{x}$ на $\frac{1}{4}$ больше значения дроби $\frac{36}{x + 20}$?
Решение:
1) $\frac{24}{x - 2} + \frac{16}{x + 2} = 3$
x − 2 ≠ 0
x ≠ 2
и
x + 2 ≠ 0
x ≠ −2
$\frac{24}{x - 2} + \frac{16}{x + 2} = 3$ | * (x − 2)(x + 2)
$24(x + 2) + 16(x - 2) = 3(x^2 - 4)$
$24x + 48 + 16x - 32 = 3x^2 - 12$
$-3x^2 + 40x + 16 + 12 = 0$
$-3x^2 + 40x + 28 = 0$
$D = b^2 - 4ac = 40^2 - 4 * (-3) * 28 = 1600 + 336 = 1936 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-40 + \sqrt{1936}}{2 * (-3)} = \frac{-40 + 44}{-6} = \frac{4}{-6} = -\frac{2}{3}$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-40 - \sqrt{1936}}{2 * (-3)} = \frac{-40 - 44}{-6} = \frac{-84}{-6} = 14$ − не удовлетворяет условию, так как x ≠ 10.
Ответ: при $x = -\frac{2}{3}$ и x = 14
2) $\frac{42}{x} - \frac{36}{x + 20} = \frac{1}{4}$
x ≠ 0
и
x + 20 ≠ 0
x ≠ −20
$\frac{42}{x} - \frac{36}{x + 20} = \frac{1}{4}$ | * 4x(x + 20)
42 * 4(x + 20) − 36 * 4x = x(x + 20)
$168x + 3360 - 144x = x^2 + 20x$
$-x^2 + 24x + 3360 - 20x = 0$
$-x^2 + 4x + 3360 = 0$ | * (−1)
$x^2 - 4x - 3360 = 0$
$D = b^2 - 4ac = (-4)^2 - 4 * 1 * (-3360) = 16 + 13440 = 13456 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{4 + \sqrt{13456}}{2 * 1} = \frac{4 + 116}{2} = \frac{120}{2} = 60$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{4 - \sqrt{13456}}{2 * 1} = \frac{4 - 116}{2} = \frac{-112}{2} = -56$
Ответ: при x = −56 и x = 60
790. При каком значении переменной:
1) значение дроби $\frac{30}{x + 3}$ на $\frac{1}{2}$ меньше значения дроби $\frac{30}{x}$;
2) значение дроби $\frac{20}{x}$ на 9 больше значения дроби $\frac{20}{x + 18}$?
Решение:
1) $\frac{30}{x} - \frac{30}{x + 3} = \frac{1}{2}$
x ≠ 0
и
x + 3 ≠ 0
x ≠ −3
$\frac{30}{x} - \frac{30}{x + 3} = \frac{1}{2}$ | * 2x(x + 3)
30 * 2(x + 3) − 2x * 30 = x(x + 3)
$60x + 180 - 60x = x^2 + 3x$
$-x^2 - 3x + 180 = 0$ | * (−1)
$x^2 + 3x - 180 = 0$
$D = b^2 - 4ac = 3^2 - 4 * 1 * (-180) = 9 + 720 = 729 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-3 + \sqrt{729}}{2 * 1} = \frac{-3 + 27}{2} = \frac{24}{2} = 12$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-3 - \sqrt{729}}{2 * 1} = \frac{-3 - 27}{2} = \frac{-30}{2} = -15$
Ответ: при x = −15 и x = 12
2) $\frac{20}{x} - \frac{20}{x + 18} = 9$
x ≠ 0
и
x + 18 ≠ 0
x ≠ −18
$\frac{20}{x} - \frac{20}{x + 18} = 9$ | * x(x + 18)
20(x + 18) − 20x = 9x(x + 18)
$20x + 360 - 20x = 9x^2 + 162x$
$-9x^2 -162x + 360 = 0$ | : (−9)
$x^2 + 18x - 40 = 0$
$D = b^2 - 4ac = 18^2 - 4 * 1 * (-40) = 324 + 160 = 484 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-18 + \sqrt{484}}{2 * 1} = \frac{-18 + 22}{2} = \frac{4}{2} = 2$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-18 - \sqrt{484}}{2 * 1} = \frac{-18 - 22}{2} = \frac{-40}{2} = -20$
Ответ: при x = −20 и x = 2
791. Решите уравнение:
1) $\frac{2x - 10}{x^3 + 1} + \frac{4}{x + 1} = \frac{5x - 1}{x^2 - x + 1}$;
2) $\frac{6}{x^2 - 4x + 3} + \frac{5 - 2x}{x - 1} = \frac{3}{x - 3}$;
3) $\frac{4x - 6}{x + 2} - \frac{x}{x + 1} = \frac{14}{x^2 + 3x + 2}$;
4) $\frac{x}{x^2 - 4} - \frac{3x - 1}{x^2 + x - 6} = \frac{2}{x^2 + 5x + 6}$.
Решение:
1) $\frac{2x - 10}{x^3 + 1} + \frac{4}{x + 1} = \frac{5x - 1}{x^2 - x + 1}$
$\frac{2x - 10}{(x + 1)(x^2 - x + 1)} + \frac{4}{x + 1} = \frac{5x - 1}{x^2 - x + 1}$
x + 1 ≠ 0
x ≠ −1
и
$x^2 - x + 1 ≠ 0$
$D = b^2 - 4ac = (-1)^2 - 4 * 1 * 1 = 1 - 4 = -3 < 0$ − нет корней.
$\frac{2x - 10}{(x + 1)(x^2 - x + 1)} + \frac{4}{x + 1} - \frac{5x - 1}{x^2 - x + 1} = 0$ | * $(x + 1)(x^2 - x + 1)$
$2x - 10 + 4(x^2 - x + 1) - (5x - 1)(x + 1) = 0$
$2x - 10 + 4x^2 - 4x + 4 - (5x^2 - x + 5x - 1) = 0$
$4x^2 - 2x - 6 - 5x^2 - 4x + 1 = 0$
$-x^2 - 6x - 5 = 0$ | * (−1)
$x^2 + 6x + 5 = 0$
$D = b^2 - 4ac = 6^2 - 4 * 1 * 5 = 36 - 20 = 16 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-6 + \sqrt{16}}{2 * 1} = \frac{-6 + 4}{2} = \frac{-2}{2} = -1$ − не удовлетворяет условию, так как x ≠ −1.
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-6 - \sqrt{16}}{2 * 1} = \frac{-6 - 4}{2} = \frac{-10}{2} = -5$
Ответ: −5
2) $\frac{6}{x^2 - 4x + 3} + \frac{5 - 2x}{x - 1} = \frac{3}{x - 3}$
x − 1 ≠ 0
x ≠ 1
и
x − 3 ≠ 0
x ≠ 3
и
$x^2 - 4x + 3 ≠ 0$
$D = b^2 - 4ac = (-4)^2 - 4 * 1 * 3 = 16 - 12 = 4 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{4 + \sqrt{4}}{2 * 1} = \frac{4 + 2}{2} = \frac{6}{2} ≠ 3$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{4 - \sqrt{4}}{2 * 1} = \frac{4 - 2}{2} = \frac{2}{2} ≠ 1$
$x^2 - 4x + 3 = (x - 3)(x - 1)$
$\frac{6}{(x - 3)(x - 1)} + \frac{5 - 2x}{x - 1} - \frac{3}{x - 3} = 0$ | * (x − 3)(x − 1)
$6 + (5 - 2x)(x - 3) - 3(x - 1) = 0$
$6 + 5x - 2x^2 - 15 + 6x - 3x + 3 = 0$
$-2x^2 + 8x - 6 = 0$ | : (−2)
$x^2 - 4x + 3 = 0$
$D = b^2 - 4ac = (-4)^2 - 4 * 1 * 3 = 16 - 12 = 4 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{4 + \sqrt{4}}{2 * 1} = \frac{4 + 2}{2} = \frac{6}{2} = 3$ − не удовлетворяет условию, так как x ≠ 3.
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{4 - \sqrt{4}}{2 * 1} = \frac{4 - 2}{2} = \frac{2}{2} = 1$ − не удовлетворяет условию, так как x ≠ 1.
Ответ: нет корней
3) $\frac{4x - 6}{x + 2} - \frac{x}{x + 1} = \frac{14}{x^2 + 3x + 2}$
x + 2 ≠ 0
x ≠ −2
и
x + 1 ≠ 0
x ≠ −1
и
$x^2 + 3x + 2 ≠ 0$
$D = b^2 - 4ac = 3^2 - 4 * 1 * 2 = 9 - 8 = 1 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-3 + \sqrt{1}}{2 * 1} = \frac{-3 + 1}{2} = \frac{-2}{2} ≠ -1$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-3 - \sqrt{1}}{2 * 1} = \frac{-3 - 1}{2} = \frac{-4}{2} ≠ -2$
$x^2 + 3x + 2 = (x - (-1))(x - (-2)) = (x + 1)(x + 2)$
$\frac{4x - 6}{x + 2} - \frac{x}{x + 1} - \frac{14}{(x + 1)(x + 2)} = 0$ | * (x + 1)(x + 2)
$(4x - 6)(x + 1) - x(x + 2) - 14 = 0$
$4x^2 - 6x + 4x - 6 - x^2 - 2x - 14 = 0$
$3x^2 - 4x - 20 = 0$
$D = b^2 - 4ac = (-4)^2 - 4 * 3 * (-20) = 16 + 240 = 256 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{4 + \sqrt{256}}{2 * 3} = \frac{4 + 16}{6} = \frac{20}{6} = \frac{10}{3} = 3\frac{1}{3}$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{4 - \sqrt{256}}{2 * 3} = \frac{4 - 16}{6} = \frac{-12}{6} = -4$
Ответ: −4 и $3\frac{1}{3}$
4) $\frac{x}{x^2 - 4} - \frac{3x - 1}{x^2 + x - 6} = \frac{2}{x^2 + 5x + 6}$
$x^2 - 4 ≠ 0$
(x − 2)(x + 2) ≠ 0
x − 2 ≠ 0
x ≠ 2
и
x + 2 ≠ 0
x ≠ −2
и
$x^2 + x - 6 ≠ 0$
$D = b^2 - 4ac = 1^2 - 4 * 1 * (-6) = 1 + 24 = 25 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-1 + \sqrt{25}}{2 * 1} = \frac{-1 + 5}{2} = \frac{4}{2} ≠ 2$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-1 - \sqrt{25}}{2 * 1} = \frac{-1 - 5}{2} = \frac{-6}{2} ≠ -3$
$x^2 + x - 6 = (x - 2)(x - (-3)) = (x - 2)(x + 3)$
$x^2 + 5x + 6 ≠ 0$
$D = b^2 - 4ac = 5^2 - 4 * 1 * 6 = 25 - 24 = 1 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-5 + \sqrt{1}}{2 * 1} = \frac{-5 + 1}{2} = \frac{-4}{2} ≠ -2$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-5 - \sqrt{1}}{2 * 1} = \frac{-5 - 1}{2} = \frac{-6}{2} ≠ -3$
$x^2 + 5x + 6 = (x - (-2))(x - (-3)) = (x + 2)(x + 3)$
$\frac{x}{(x - 2)(x + 2)} - \frac{3x - 1}{(x - 2)(x + 3)} - \frac{2}{(x + 2)(x + 3)} = 0$ | * (x − 2)(x + 2)(x + 3)
x(x + 3) − (3x − 1)(x + 2) − 2(x − 2) = 0
$x^2 + 3x - (3x^2 - x + 6x - 2) - 2x + 4 = 0$
$x^2 + 3x - 3x^2 + x - 6x + 2 - 2x + 4 = 0$
$-2x^2 - 4x + 6 = 0$ | : (−2)
$x^2 + 2x - 3 = 0$
$D = b^2 - 4ac = 2^2 - 4 * 1 * (-3) = 4 + 12 = 16 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-2 + \sqrt{16}}{2 * 1} = \frac{-2 + 4}{2} = \frac{2}{2} = 1$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-2 - \sqrt{16}}{2 * 1} = \frac{-2 - 4}{2} = \frac{-6}{2} = -3$ − не удовлетворяет условию, так как x ≠ −3.
Ответ: 1
792. Решите уравнение:
1) $\frac{3x + 2}{x^2 + 2x + 4} + \frac{x^2 + 39}{x^3 - 8} = \frac{5}{x - 2}$;
2) $\frac{x}{x - 1} + \frac{x + 1}{x + 3} = \frac{8}{x^2 + 2x - 3}$.
Решение:
1) $\frac{3x + 2}{x^2 + 2x + 4} + \frac{x^2 + 39}{x^3 - 8} = \frac{5}{x - 2}$
$\frac{3x + 2}{x^2 + 2x + 4} + \frac{x^2 + 39}{(x - 2)(x^2 + 2x + 4)} = \frac{5}{x - 2}$
x − 2 ≠ 0
x ≠ 2
и
$x^2 + 2x + 4 ≠ 0$
$D = b^2 - 4ac = 2^2 - 4 * 1 * 4 = 4 - 16 = -12 < 0$ − нет корней
$\frac{3x + 2}{x^2 + 2x + 4} + \frac{x^2 + 39}{(x - 2)(x^2 + 2x + 4)} - \frac{5}{x - 2} = 0$ | * $(x - 2)(x^2 + 2x + 4)$
$(3x + 2)(x - 2) + x^2 + 39 - 5(x^2 + 2x + 4) = 0$
$3x^2 + 2x - 6x - 4 + x^2 + 39 - 5x^2 - 10x - 20 = 0$
$-x^2 - 14x + 15 = 0$ | * (−1)
$x^2 + 14x - 15 = 0$
$D = b^2 - 4ac = 14^2 - 4 * 1 * (-15) = 196 + 60 = 256 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-14 + \sqrt{256}}{2 * 1} = \frac{-14 + 16}{2} = \frac{2}{2} = 1$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-14 - \sqrt{256}}{2 * 1} = \frac{-14 - 16}{2} = \frac{-30}{2} = -15$
Ответ: −15 и 1
2) $\frac{x}{x - 1} + \frac{x + 1}{x + 3} = \frac{8}{x^2 + 2x - 3}$
x − 1 ≠ 0
x ≠ 1
и
x + 3 ≠ 0
x ≠ −3
и
$x^2 + 2x - 3 ≠ 0$
$D = b^2 - 4ac = 2^2 - 4 * 1 * (-3) = 4 + 12 = 16 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-2 + \sqrt{16}}{2 * 1} = \frac{-2 + 4}{2} = \frac{2}{2} ≠ 1$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-2 - \sqrt{16}}{2 * 1} = \frac{-2 - 4}{2} = \frac{-6}{2} ≠ -3$
$x^2 + 2x - 3 = (x - 1)(x - (-3)) = (x - 1)(x + 3)$
$\frac{x}{x - 1} + \frac{x + 1}{x + 3} - \frac{8}{(x - 1)(x + 3)} = 0$ | * (x − 1)(x + 3)
x(x + 3) + (x + 1)(x − 1) − 8 = 0
$x^2 + 3x + x^2 + x - x - 1 - 8 = 0$
$2x^2 + 3x - 9 = 0$
$D = b^2 - 4ac = 3^2 - 4 * 2 * (-9) = 9 + 72 = 81 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-3 + \sqrt{81}}{2 * 2} = \frac{-3 + 9}{4} = \frac{6}{4} = 1,5$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-3 - \sqrt{81}}{2 * 2} = \frac{-3 - 9}{4} = \frac{-12}{4} = -3$ − не удовлетворяет условию, так как x ≠ −3.
Ответ: 1,5
793. Решите уравнение методом замены переменной:
1) $(x^2 - 2)^2 - 8(x^2 - 2) + 7 = 0$;
2) $(x^2 + 5x)^2 - 2(x^2 + 5x) - 24 = 0$;
3) $(x^2 - 3x + 1)(x^2 - 3x + 3) = 3$;
4) $(x^2 + 2x + 2)(x^2 + 2x - 4) = -5$.
Решение:
1) $(x^2 - 2)^2 - 8(x^2 - 2) + 7 = 0$
$y = x^2 - 2$
$y^2 - 8y + 7 = 0$
$D = b^2 - 4ac = (-8)^2 - 4 * 1 * 7 = 64 - 28 = 36 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{8 + \sqrt{36}}{2 * 1} = \frac{8 + 6}{2} = \frac{14}{2} = 7$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{8 - \sqrt{36}}{2 * 1} = \frac{8 - 6}{2} = \frac{2}{2} = 1$
$x^2 - 2 = 7$
$x^2 = 7 + 2$
$x^2 = 9$
x = ±3
или
$x^2 - 2 = 1$
$x^2 = 1 + 2$
$x^2 = 3$
$x = ±\sqrt{3}$
Ответ: $-3, -\sqrt{3}, \sqrt{3}, 3.$
2) $(x^2 + 5x)^2 - 2(x^2 + 5x) - 24 = 0$
$y = x^2 + 5x$
$y^2 - 2y - 24 = 0$
$D = b^2 - 4ac = (-2)^2 - 4 * 1 * (-24) = 4 + 96 = 100 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{2 + \sqrt{100}}{2 * 1} = \frac{2 + 10}{2} = \frac{12}{2} = 6$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{2 - \sqrt{100}}{2 * 1} = \frac{2 - 10}{2} = \frac{-8}{2} = -4$
$x^2 + 5x = 6$
$x^2 + 5x - 6 = 0$
$D = b^2 - 4ac = 5^2 - 4 * 1 * (-6) = 25 + 24 = 49 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-5 + \sqrt{49}}{2 * 1} = \frac{-5 + 7}{2} = \frac{2}{2} = 1$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-5 - \sqrt{49}}{2 * 1} = \frac{-5 - 7}{2} = \frac{-12}{2} = -6$
или
$x^2 + 5x = -4$
$x^2 + 5x + 4 = 0$
$D = b^2 - 4ac = 5^2 - 4 * 1 * 4 = 25 - 16 = 9 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-5 + \sqrt{9}}{2 * 1} = \frac{-5 + 3}{2} = \frac{-2}{2} = -1$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-5 - \sqrt{9}}{2 * 1} = \frac{-5 - 3}{2} = \frac{-8}{2} = -4$
Ответ: −6, −4, −1, 1.
3) $(x^2 - 3x + 1)(x^2 - 3x + 3) = 3$
$y = x^2 - 3x$
(y + 1)(y + 3) = 3
$y^2 + y + 3y + 3 - 3 = 0$
$y^2 + 4y = 0$
y(y + 4) = 0
y = 0
или
y + 4 = 0
y = −4
$x^2 - 3x = 0$
x(x − 3) = 0
x = 0
или
x − 3 = 0
x = 3
или
$x^2 - 3x = -4$
$x^2 - 3x + 4 = 0$
$D = b^2 - 4ac = (-3)^2 - 4 * 1 * 4 = 9 - 16 = -7 < 0$ − нет корней
Ответ: 0 и 3
4) $(x^2 + 2x + 2)(x^2 + 2x - 4) = -5$
$y = x^2 + 2x$
(y + 2)(y − 4) = −5
$y^2 + 2y - 4y - 8 + 5 = 0$
$y^2 - 2y - 3 = 0$
$D = b^2 - 4ac = (-2)^2 - 4 * 1 * (-3) = 4 + 12 = 16 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{2 + \sqrt{16}}{2 * 1} = \frac{2 + 4}{2} = \frac{6}{2} = 3$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{2 - \sqrt{16}}{2 * 1} = \frac{2 - 4}{2} = \frac{-2}{2} = -1$
$x^2 + 2x = 3$
$x^2 + 2x - 3 = 0$
$D = b^2 - 4ac = 2^2 - 4 * 1 * (-3) = 4 + 12 = 16 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-2 + \sqrt{16}}{2 * 1} = \frac{-2 + 4}{2} = \frac{2}{2} = 1$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-2 - \sqrt{16}}{2 * 1} = \frac{-2 - 4}{2} = \frac{-6}{2} = -3$
или
$x^2 + 2x = -1$
$x^2 + 2x + 1 = 0$
$D = b^2 - 4ac = 2^2 - 4 * 1 * 1 = 4 - 4 = 0$
$x = \frac{-b + \sqrt{D}}{2a} = \frac{-2 + \sqrt{0}}{2 * 1} = \frac{-2}{2} = -1$
Ответ: −3, −1, 1.
794. Решите уравнение методом замены переменной:
1) $(\frac{2x - 1}{x})^2 - \frac{6(2x - 1)}{x} + 5 = 0$;
2) $\frac{3x - 1}{x + 1} + \frac{x + 1}{3x - 1} = 3\frac{1}{3}$.
Решение:
1) $(\frac{2x - 1}{x})^2 - \frac{6(2x - 1)}{x} + 5 = 0$
$y = \frac{2x - 1}{x}$
$y^2 - 6y + 5 = 0$
$D = b^2 - 4ac = (-6)^2 - 4 * 1 * 5 = 36 - 20 = 16 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{6 + \sqrt{16}}{2 * 1} = \frac{6 + 4}{2} = \frac{10}{2} = 5$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{6 - \sqrt{16}}{2 * 1} = \frac{6 - 4}{2} = \frac{2}{2} = 1$
$\frac{2x - 1}{x} = 5$
2x − 1 = 5x
2x − 5x = 1
−3x = 1
$x = -\frac{1}{3}$
или
$\frac{2x - 1}{x} = 1$
2x − 1 = x
2x − x = 1
x = 1
Ответ: $-\frac{1}{3}$ и 1
2) $\frac{3x - 1}{x + 1} + \frac{x + 1}{3x - 1} = 3\frac{1}{3}$
$\frac{3x - 1}{x + 1} + \frac{x + 1}{3x - 1} = \frac{10}{3}$
x + 1 ≠ 0
x ≠ −1
и
3x − 1 ≠ 0
3x ≠ 1
$x ≠ \frac{1}{3}$
$y = \frac{3x - 1}{x + 1}$
$y + \frac{1}{y} = \frac{10}{3}$
y ≠ 0
$y + \frac{1}{y} = \frac{10}{3}$ | * 3y
$3y^2 + 3 = 10y$
$3y^2 - 10y + 3 = 0$
$D = b^2 - 4ac = (-10)^2 - 4 * 3 * 3 = 100 - 36 = 64 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{10 + \sqrt{64}}{2 * 3} = \frac{10 + 8}{6} = \frac{18}{6} = 3$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{10 - \sqrt{64}}{2 * 3} = \frac{10 - 8}{6} = \frac{2}{6} = \frac{1}{3}$
$\frac{3x - 1}{x + 1} = 3$ | * (x + 1)
3x − 1 = 3(x + 1)
3x − 1 = 3x + 3
3x − 3x = 3 + 1
0 ≠ 4
нет корней
$\frac{3x - 1}{x + 1} = \frac{1}{3}$ | * 3(x + 1)
3(3x − 1) = x + 1
9x − 3 = x + 1
9x − x = 1 + 3
8x = 4
x = 0,5
Ответ: 0,5
795. Решите уравнение:
1) $(x^2 - 6x)^2 + (x^2 - 6x) - 56 = 0$;
2) $(x^2 + 8x + 3)(x^2 + 8x + 5) = 63$;
3) $\frac{x^4}{(x - 2)^2} - \frac{4x^2}{x - 2} - 5 = 0$;
4) $\frac{x + 4}{x - 3} - \frac{x - 3}{x + 4} = \frac{3}{2}$.
Решение:
1) $(x^2 - 6x)^2 + (x^2 - 6x) - 56 = 0$
$y = x^2 - 6x$
$y^2 + y - 56 = 0$
$D = b^2 - 4ac =1^2 - 4 * 1 * (-56) = 1 + 224 = 225 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-1 + \sqrt{225}}{2 * 1} = \frac{-1 + 15}{2} = \frac{14}{2} = 7$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-1 - \sqrt{225}}{2 * 1} = \frac{-1 - 15}{2} = \frac{-16}{2} = -8$
$x^2 - 6x = 7$
$x^2 - 6x - 7 = 0$
$D = b^2 - 4ac =(-6)^2 - 4 * 1 * (-7) = 36 + 28 = 64 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{6 + \sqrt{64}}{2 * 1} = \frac{6 + 8}{2} = \frac{14}{2} = 7$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{6 - \sqrt{64}}{2 * 1} = \frac{6 - 8}{2} = \frac{-2}{2} = -1$
или
$x^2 - 6x = -8$
$x^2 - 6x + 8 = 0$
$D = b^2 - 4ac =(-6)^2 - 4 * 1 * 8 = 36 - 32 = 4 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{6 + \sqrt{4}}{2 * 1} = \frac{6 + 2}{2} = \frac{8}{2} = 4$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{6 - \sqrt{4}}{2 * 1} = \frac{6 - 2}{2} = \frac{4}{2} = 2$
Ответ: −1, 2, 4, 7.
2) $(x^2 + 8x + 3)(x^2 + 8x + 5) = 63$
$y = x^2 + 8x$
$(y + 3)(y + 5) = 63$
$y^2 + 3y + 5y + 15 - 63 = 0$
$y^2 + 8y - 48 = 0$
$D = b^2 - 4ac =8^2 - 4 * 1 * (-48) = 64 + 192 = 256 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-8 + \sqrt{256}}{2 * 1} = \frac{-8 + 16}{2} = \frac{8}{2} = 4$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-8 - \sqrt{256}}{2 * 1} = \frac{-8 - 16}{2} = \frac{-24}{2} = -12$
$x^2 + 8x = 4$
$x^2 + 8x - 4 = 0$
$D = b^2 - 4ac =8^2 - 4 * 1 * (-4) = 64 + 16 = 80 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-8 + \sqrt{80}}{2 * 1} = \frac{-8 + \sqrt{16 * 5}}{2} = \frac{-8 + 4\sqrt{5}}{2} = -4 + 2\sqrt{5}$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-8 - \sqrt{80}}{2 * 1} = \frac{-8 - \sqrt{16 * 5}}{2} = \frac{-8 - 4\sqrt{5}}{2} = -4 - 2\sqrt{5}$
или
$x^2 + 8x = -12$
$x^2 + 8x + 12 = 0$
$D = b^2 - 4ac =8^2 - 4 * 1 * 12 = 64 - 48 = 16 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-8 + \sqrt{16}}{2 * 1} = \frac{-8 + 4}{2} = \frac{-4}{2} = -2$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-8 - \sqrt{16}}{2 * 1} = \frac{-8 - 4}{2} = \frac{-12}{2} = -6$
Ответ: $-4 - 2\sqrt{5}, -4 + 2\sqrt{5}, -6, -2.$
3) $\frac{x^4}{(x - 2)^2} - \frac{4x^2}{x - 2} - 5 = 0$
$(x - 2)^2 ≠ 0$
x − 2 ≠ 0
x ≠ 2
$(\frac{x^2}{x - 2})^2 - \frac{4x^2}{x - 2} - 5 = 0$
$y = \frac{x^2}{x - 2}$
$y^2 - 4y - 5 = 0$
$D = b^2 - 4ac =(-4)^2 - 4 * 1 * (-5) = 16 + 20 = 36 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{4 + \sqrt{36}}{2 * 1} = \frac{4 + 6}{2} = \frac{10}{2} = 5$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{4 - \sqrt{36}}{2 * 1} = \frac{4 - 6}{2} = \frac{-2}{2} = -1$
$\frac{x^2}{x - 2} = 5$ | * (x − 2)
$x^2 = 5(x - 2)$
$x^2 = 5x - 10$
$x^2 - 5x + 10 = 0$
$D = b^2 - 4ac =(-5)^2 - 4 * 1 * 10 = 25 - 40 = 15 < 0$ − нет корней
или
$\frac{x^2}{x - 2} = -1$ | * (x − 2)
$x^2 = -(x - 2)$
$x^2 = -x + 2$
$x^2 + x - 2 = 0$
$D = b^2 - 4ac =1^2 - 4 * 1 * (-2) = 1 + 8 = 9 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-1 + \sqrt{9}}{2 * 1} = \frac{-1 + 3}{2} = \frac{2}{2} = 1$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-1 - \sqrt{9}}{2 * 1} = \frac{-1 - 3}{2} = \frac{-4}{2} = -2$
Ответ: −2 и 1
4) $\frac{x + 4}{x - 3} - \frac{x - 3}{x + 4} = \frac{3}{2}$
x − 3 ≠ 0
x ≠ 3
и
x + 4 ≠ 0
x ≠ −4
$y = \frac{x + 4}{x - 3}$
$y - \frac{1}{y} - \frac{3}{2} = 0$ | * 2y
$2y^2 - 3y - 2 = 0$
$D = b^2 - 4ac =(-3)^2 - 4 * 2 * (-2) = 9 + 16 = 25 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{3 + \sqrt{25}}{2 * 2} = \frac{3 + 5}{4} = \frac{8}{4} = 2$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{3 - \sqrt{25}}{2 * 2} = \frac{3 - 5}{4} = \frac{-2}{4} = -\frac{1}{2}$
$\frac{x + 4}{x - 3} = 2$
x + 4 = 2(x − 3)
x + 4 = 2x − 6
x − 2x = −6 − 4
−x = −10
x = 10
или
$\frac{x + 4}{x - 3} = -\frac{1}{2}$ | * 2(x − 3)
2(x + 4) = −(x − 3)
2x + 8 = −x + 3
2x + x = 3 − 8
3x = −5
$x = -\frac{5}{3} = -1\frac{2}{3}$
Ответ: $-1\frac{2}{3}$ и 10
