Ответы к странице 196
Когда сделаны уроки
1. Решите уравнение:
1) $\frac{3x^2 - 9x}{2} - \frac{12}{x^2 - 3x} = 3$;
2) $\frac{6}{(x + 1)(x + 2)} + \frac{8}{(x - 1)(x + 4)} = 1$;
3) x(x + 3)(x + 5)(x + 8) = 100;
4) $(x + 2)(x + 3)(x + 8)(x + 12) = 4x^2$;
5) $7(x + \frac{1}{x}) - 2(x^2 + \frac{1}{x^2}) = 9$;
6) $2(x^2 + x + 1)^2 - 7(x - 1)^2 = 13(x^3 - 1)$;
7) $(x - 6)^4 + (x - 4)^4 = 82$.
Решение:
1) $\frac{3x^2 - 9x}{2} - \frac{12}{x^2 - 3x} = 3$
$\frac{3(x^2 - 3x)}{2} - \frac{12}{x^2 - 3x} = 3$
$x^2 - 3x ≠ 0$
x(x − 3) ≠ 0
x ≠ 0
или
x − 3 ≠ 0
x ≠ 3
$y = x^2 - 3x$
$\frac{3y}{2} - \frac{12}{y} = 3$ | * 2y
$3y^2 - 24 = 6y$
$3y^2 - 6y - 24 = 0$ | : 3
$y^2 - 2y - 8 = 0$
$D = b^2 - 4ac =(-2)^2 - 4 * 1 * (-8) = 4 + 32 = 36 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{2 + \sqrt{36}}{2 * 1} = \frac{2 + 6}{2} = \frac{8}{2} = 4$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{2 - \sqrt{36}}{2 * 1} = \frac{2 - 6}{2} = \frac{-4}{2} = -2$
$x^2 - 3x = 4$
$x^2 - 3x - 4 = 0$
$D = b^2 - 4ac =(-3)^2 - 4 * 1 * (-4) = 9 + 16 = 25 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{3 + \sqrt{25}}{2 * 1} = \frac{3 + 5}{2} = \frac{8}{2} = 4$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{3 - \sqrt{25}}{2 * 1} = \frac{3 - 5}{2} = \frac{-2}{2} = -1$
или
$x^2 - 3x = -2$
$x^2 - 3x + 2 = 0$
$D = b^2 - 4ac =(-3)^2 - 4 * 1 * 2 = 9 - 8 = 1 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{3 + \sqrt{1}}{2 * 1} = \frac{3 + 1}{2} = \frac{4}{2} = 2$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{3 - \sqrt{1}}{2 * 1} = \frac{3 - 1}{2} = \frac{2}{2} = 1$
Ответ: −1, 1, 2, 4.
2) $\frac{6}{(x + 1)(x + 2)} + \frac{8}{(x - 1)(x + 4)} = 1$
x + 1 ≠ 0
x ≠ −1
и
x + 2 ≠ 0
x ≠ −2
и
x − 1 ≠ 0
x ≠ 1
и
x + 4 ≠ 0
x ≠ −4
$\frac{6}{x^2 + x + 2x + 2} + \frac{8}{x^2 - x + 4x - 4} = 1$
$\frac{6}{x^2 + 3x + 2} + \frac{8}{x^2 + 3x - 4} = 1$
$y = x^2 + 3x$
$\frac{6}{y + 2} + \frac{8}{y - 4} = 1$
y + 2 ≠ 0
y ≠ −2
и
y − 4 ≠ 0
y ≠ 4
$\frac{6}{y + 2} + \frac{8}{y - 4} - 1 = 0$ | * (y + 2)(y − 4)
6(y − 4) + 8(y + 2) − (y + 2)(y − 4) = 0
$6y - 24 + 8y + 16 - (y^2 + 2y - 4y - 8) = 0$
$14y - 8 - y^2 + 2y + 8 = 0$
$-y^2 + 16y = 0$ | * (−1)
$y^2 - 16y = 0$
$y(y - 16) = 0$
y = 0
или
y − 16 = 0
y = 16
$x^2 + 3x = 0$
x(x + 3) = 0
x = 0
или
x + 3 = 0
x = −3
или
$x^2 + 3x = 16$
$x^2 + 3x - 16 = 0$
$D = b^2 - 4ac =(-3)^2 - 4 * 1 * (-16) = 9 + 64 = 73 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-3 + \sqrt{73}}{2 * 1} = \frac{-3 + \sqrt{73}}{2}$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-3 - \sqrt{73}}{2 * 1} = \frac{-3 - \sqrt{73}}{2}$
Ответ: $-3, 0, \frac{-3 + \sqrt{73}}{2}, \frac{-3 - \sqrt{73}}{2}$.
3) x(x + 3)(x + 5)(x + 8) = 100
((x + 8))((x + 3)(x + 5)) = 100
$(x^2 + 8x)(x^2 + 3x + 5x + 15) = 100$
$(x^2 + 8x)(x^2 + 8x + 15) = 100$
$y = x^2 + 8x$
y(y + 15) = 100
$y^2 + 15y - 100 = 0$
$D = b^2 - 4ac =15^2 - 4 * 1 * (-100) = 225 + 400 = 625 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-15 + \sqrt{625}}{2 * 1} = \frac{-15 + 25}{2} = \frac{10}{2} = 5$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-15 - \sqrt{625}}{2 * 1} = \frac{-15 - 25}{2} = \frac{-40}{2} = -20$
$x^2 + 8x = 5$
$x^2 + 8x - 5 = 0$
$D = b^2 - 4ac = 8^2 - 4 * 1 * (-5) = 64 + 20 = 84 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-8 + \sqrt{84}}{2 * 1} = \frac{-8 + \sqrt{4 * 21}}{2} = \frac{-8 + 2\sqrt{21}}{2} = \frac{2(-4 + \sqrt{21})}{2} = -4 + \sqrt{21}$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-8 - \sqrt{84}}{2 * 1} = \frac{-8 - \sqrt{4 * 21}}{2} = \frac{-8 - 2\sqrt{21}}{2} = \frac{2(-4 - \sqrt{21})}{2} = -4 - \sqrt{21}$
или
$x^2 + 8x = -20$
$x^2 + 8x + 20 = 0$
$D = b^2 - 4ac = 8^2 - 4 * 1 *20 = 64 - 80 = -16 < 0$ − нет корней
Ответ: $-4 + \sqrt{21}$ и $-4 - \sqrt{21}$
4) $(x + 2)(x + 3)(x + 8)(x + 12) = 4x^2$
$((x + 2)(x + 12))((x + 3)(x + 8)) = 4x^2$
$(x^2 + 2x + 12x + 24)(x^2 + 3x + 8x + 24) = 4x^2$
$(x^2 + 14x + 24)(x^2 + 11x + 24) = 4x^2$
x = 0 не является корнем уравнения, обе части уравнения разделим на $x^2$.
$\frac{(x^2 + 14x + 24)(x^2 + 11x + 24)}{x^2} = \frac{4x^2}{x^2}$
$\frac{x^2 + 14x + 24}{x} * \frac{x^2 + 11x + 24}{x} = 4$
$(x + 14 + \frac{24}{x})(x + 11 + \frac{24}{x}) = 4$
$(x + \frac{24}{x} + 14)(x + \frac{24}{x} + 11) = 4$
$y = x + \frac{24}{x}$
(y + 14)(y + 11) = 4
$y^2 + 14y + 11y + 154 - 4 = 0$
$y^2 + 25y + 150 = 0$
$D = b^2 - 4ac =25^2 - 4 * 1 * 150 = 625 - 600 = 25 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-25 + \sqrt{25}}{2 * 1} = \frac{-25 + 5}{2} = \frac{-20}{2} = -10$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-25 - \sqrt{25}}{2 * 1} = \frac{-25 - 5}{2} = \frac{-30}{2} = -15$
$x + \frac{24}{x} = -10$ | * x
$x^2 + 24 = -10x$
$x^2 + 10x + 24 = 0$
$D = b^2 - 4ac =10^2 - 4 * 1 * 24 = 100 - 96 = 4 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-10 + \sqrt{4}}{2 * 1} = \frac{-10 + 2}{2} = \frac{-8}{2} = -4$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-10 - \sqrt{4}}{2 * 1} = \frac{-10 - 2}{2} = \frac{-12}{2} = -6$
или
$x + \frac{24}{x} = -15$ | * x
$x^2 + 24 = -15x$
$x^2 + 15x + 24 = 0$
$D = b^2 - 4ac =15^2 - 4 * 1 * 24 = 225 - 96 = 129 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-10 + \sqrt{129}}{2 * 1} = \frac{-10 + \sqrt{129}}{2}$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-10 - \sqrt{129}}{2 * 1} = \frac{-10 - \sqrt{129}}{2}$
Ответ: $-6, -4, \frac{-10 + \sqrt{129}}{2}, \frac{-10 - \sqrt{129}}{2}$.
5) $7(x + \frac{1}{x}) - 2(x^2 + \frac{1}{x^2}) = 9$
x ≠ 0
$y = x + \frac{1}{x}$
$y^2 = (x + \frac{1}{x})^2 = x^2 + 2 * x * \frac{1}{x} + (\frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2}$
тогда:
$x^2 + \frac{1}{x^2} = y^2 - 2$
$7y - 2(y^2 - 2) = 9$
$7y - 2y^2 + 4 - 9 = 0$
$-2y^2 + 7y - 5 = 0$ | * (−1)
$2y^2 - 7y + 5 = 0$
$D = b^2 - 4ac = (-7)^2 - 4 * 2 * 5 = 49 - 40 = 9 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{7 + \sqrt{9}}{2 * 2} = \frac{7 + 3}{4} = \frac{10}{4} = \frac{5}{2} = 2,5$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{7 - \sqrt{9}}{2 * 2} = \frac{7 - 3}{4} = \frac{4}{4} = 1$
1)
y = 2,5
$x + \frac{1}{x} = 2,5$ | * 2x
$2x^2 + 2 = 5x$
$2x^2 - 5x + 2 = 0$
$D = b^2 - 4ac =(-5)^2 - 4 * 2 * 2 = 25 - 16 = 9 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{5 + \sqrt{9}}{2 * 2} = \frac{5 + 3}{4} = \frac{8}{4} = 2$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{5 - \sqrt{9}}{2 * 2} = \frac{5 - 3}{4} = \frac{2}{4} = \frac{1}{2} = 0,5$
2)
y = 1
$x + \frac{1}{x} = 1$ | * x
$x^2 + 1 = x$
$x^2 - x + 1 = 0$
$D = b^2 - 4ac =(-1)^2 - 4 * 1 * 1 = 1 - 4 = -3 < 0$ − нет корней
Ответ: 0,5 и 2
6) $2(x^2 + x + 1)^2 - 7(x - 1)^2 = 13(x^3 - 1)$
$2(x^2 + x + 1)^2 - 7(x - 1)^2 = 13(x - 1)(x^2 + x + 1)$
$x^3 - 1 ≠ 0$
x ≠ 1
$2(x^2 + x + 1)^2 - 7(x - 1)^2 = 13(x - 1)(x^2 + x + 1)$ | : $(x - 1)(x^2 + x + 1)$
$\frac{2(x^2 + x + 1)^2}{(x - 1)(x^2 + x + 1)} - \frac{7(x - 1)^2}{(x - 1)(x^2 + x + 1)} = \frac{13(x - 1)(x^2 + x + 1)}{(x - 1)(x^2 + x + 1)}$
$\frac{2(x^2 + x + 1)}{x - 1} - \frac{7(x - 1)}{x^2 + x + 1} = 13$
$y = \frac{x^2 + x + 1}{x - 1}$
$2y - \frac{7}{y} = 13$ | * y
$2y^2 - 7 = 13y$
$2y^2 - 13y - 7 = 0$
$D = b^2 - 4ac = (-13)^2 - 4 * 2 * (-7) = 169 + 56 = 225 > 0$
$y_1 = \frac{-b + \sqrt{D}}{2a} = \frac{13 + \sqrt{225}}{2 * 2} = \frac{13 + 15}{4} = \frac{28}{4} = 7$
$y_2 = \frac{-b - \sqrt{D}}{2a} = \frac{13 - \sqrt{225}}{2 * 2} = \frac{13 - 15}{4} = \frac{-2}{4} = -0,5$
1)
y = 7:
$\frac{x^2 + x + 1}{x - 1} = 7$ | * (x − 1)
$x^2 + x + 1 = 7(x - 1)$
$x^2 + x + 1 = 7x - 7$
$x^2 + x - 7x + 1 + 7 = 0$
$x^2 - 6x + 8 = 0$
$D = b^2 - 4ac =(-6)^2 - 4 * 1 * 8 = 36 - 32 = 4 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{6 + \sqrt{4}}{2 * 1} = \frac{6 + 2}{2} = \frac{8}{2} = 4$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{6 - \sqrt{4}}{2 * 1} = \frac{6 - 2}{2} = \frac{4}{2} = 2$
2)
y = −0,5:
$\frac{x^2 + x + 1}{x - 1} = -0,5$ | * 2(x − 1)
$2(x^2 + x + 1) = -(x - 1)$
$2x^2 + 2x + 2 = -x + 1$
$2x^2 + 2x + 2 + x - 1 = 0$
$2x^2 + 3x + 1 = 0$
$D = b^2 - 4ac =3^2 - 4 * 2 * 1 = 9 - 8 = 1 > 0$
$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-3 + \sqrt{1}}{2 * 2} = \frac{-3 + 1}{4} = \frac{-2}{4} = -0,5$
$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-3 - \sqrt{1}}{2 * 2} = \frac{-3 - 1}{4} = \frac{-4}{4} = -1$
Ответ: −1, −0,5, 2, 4.
7) $(x - 6)^4 + (x - 4)^4 = 82$
$((x - 6)^2)^2 + ((x - 4)^2)^2 = 82$
сумма квадратов двух неотрицательных чисел может быть равна 82, если эти числа равны 1 и 81, тогда:
1)
$(x - 6)^4 = 1$
$((x - 6)^2)^2 = 1$
$(x - 6)^2 = 1$
$x_1 = 7$
$x_2 = 5$ − не является решением, так как:
$(5 - 6)^4 + (5 - 4)^4 = 82$
$1^4 + 1^4 = 82$
1 + 1 = 82
2 ≠ 82
или
$(x - 6)^4 = -1$ − нет корней
$(x - 4)^4 = 81$
$(x - 4)^2 = 9$
$x_1 = 7$
$x_2 = 1$ − не является решением, так как:
$(1 - 6)^4 + (1 - 4)^4 = 82$
$(-5)^4 + (-3)^4 = 82$
625 + 81 = 82
706 ≠ 82
или
$(x - 4)^2 = -9$ − нет корней
2)
$(x - 6)^4 = 81$
$(x - 6)^2 = 9$
$x_1 = 3$
$x_2 = 9$ − не является решением, так как:
$(9 - 6)^4 + (9 - 4)^4 = 82$
$3^4 + 5^4 = 82$
81 + 625 = 82
706 ≠ 82
или
$(x - 6)^2 = -9$ − нет корней
$(x - 4)^4 = 1$
$((x - 4)^2)^2 = 1$
$(x - 4)^2 = 1$
$x_1 = 3$
$x_2 = 5$ − не является решением, так как:
$(5 - 6)^4 + (5 - 4)^4 = 82$
$1^4 + 1^4 = 82$
1 + 1 = 82
2 ≠ 82
или
$(x - 4)^4 = -1$ − нет корней
Ответ: 3 и 7.
