Ответы к странице 29
113. Упростите выражение:
1) $\frac{m + n}{m - n} - \frac{m^2 + n^2}{m^2 - n^2}$;
2) $\frac{x - y}{x + y} + \frac{y^2}{2xy + x^2 + y^2}$;
3) $\frac{2a}{4a^2 - 1} - \frac{a + 4}{2a^2 + a}$;
4) $\frac{b - 2}{b^2 + 6b + 9} - \frac{b}{b^2 - 9}$;
5) $\frac{x - 6}{x^2 + 3x} + \frac{x}{x + 3} - \frac{x - 3}{x}$;
6) $\frac{y + 2}{y - 2} - \frac{y - 2}{y + 2} - \frac{16}{y^2 - 4}$.
Решение:
1) $\frac{m + n}{m - n} - \frac{m^2 + n^2}{m^2 - n^2} = \frac{m + n}{m - n} - \frac{m^2 + n^2}{(m - n)(m + n)} = \frac{(m + n)(m + n) - (m^2 + n^2)}{(m - n)(m + n)} = \frac{(m + n)^2 - (m^2 + n^2)}{(m - n)(m + n)} = \frac{m^2 + 2mn + n^2 - m^2 - n^2}{(m - n)(m + n)} = \frac{2mn}{m^2 - n^2}$
2) $\frac{x - y}{x + y} + \frac{y^2}{2xy + x^2 + y^2} = \frac{x - y}{x + y} + \frac{y^2}{(x + y)^2} = \frac{(x - y)(x + y) + y^2}{(x + y)^2} = \frac{x^2 - y^2 + y^2}{(x + y)^2} = \frac{x^2}{(x + y)^2}$
3) $\frac{2a}{4a^2 - 1} - \frac{a + 4}{2a^2 + a} = \frac{2a}{(2a - 1)(2a + 1)} - \frac{a + 4}{a(2a + 1)} = \frac{2a * a - (2a - 1)(a + 4)}{a(2a - 1)(2a + 1)} = \frac{2a^2 - (2a^2 - a + 8a - 4)}{a(2a - 1)(2a + 1)} = \frac{2a^2 - 2a^2 + a - 8a + 4}{a(2a - 1)(2a + 1)} = \frac{4 - 7a}{a(4a^2 - 1)}$
4) $\frac{b - 2}{b^2 + 6b + 9} - \frac{b}{b^2 - 9} = \frac{b - 2}{(b + 3)^2} - \frac{b}{(b - 3)(b + 3)} = \frac{(b - 2)(b - 3) - b(b + 3)}{(b + 3)^2(b - 3)} = \frac{b^2 - 2b - 3b + 6 - b^2 - 3b}{(b + 3)^2(b - 3)} = \frac{6 - 8b}{(b + 3)^2(b - 3)}$
5) $\frac{x - 6}{x^2 + 3x} + \frac{x}{x + 3} - \frac{x - 3}{x} = \frac{x - 6}{x(x + 3)} + \frac{x}{x + 3} - \frac{x - 3}{x} = \frac{x - 6 + x * x - (x - 3)(x + 3)}{x(x + 3)} = \frac{x - 6 + x^2 - (x^2 - 9)}{x(x + 3)} = \frac{x - 6 + x^2 - x^2 + 9}{x(x + 3)} = \frac{x + 3}{x(x + 3)} = \frac{1}{x}$
6) $\frac{y + 2}{y - 2} - \frac{y - 2}{y + 2} - \frac{16}{y^2 - 4} = \frac{y + 2}{y - 2} - \frac{y - 2}{y + 2} - \frac{16}{(y - 2)(y + 2)} = \frac{(y + 2)(y + 2) - (y - 2)(y - 2) - 16}{(y - 2)(y + 2)} = \frac{(y + 2)^2 - (y - 2)^2 - 16}{(y - 2)(y + 2)} = \frac{y^2 + 4y + 4 - (y^2 - 4y + 4) - 16}{(y - 2)(y + 2)} = \frac{y^2 + 4y + 4 - y^2 + 4y - 4 - 16}{(y - 2)(y + 2)} = \frac{8y - 16}{(y - 2)(y + 2)} = \frac{8(y - 2)}{(y - 2)(y + 2)} = \frac{8}{y + 2}$
114. Докажите, что при всех допустимых значениях переменной значение выражения не зависит от значения переменной:
1) $\frac{2x + 1}{2x - 4} + \frac{2x - 1}{6 - 3x} - \frac{x + 7}{6x - 12}$;
2) $\frac{24 - 2a}{a^2 - 16} - \frac{a}{2a - 8} + \frac{4}{a + 4}$.
Решение:
1) $\frac{2x + 1}{2x - 4} + \frac{2x - 1}{6 - 3x} - \frac{x + 7}{6x - 12} = \frac{2x + 1}{2x - 4} - \frac{2x - 1}{3x - 6} - \frac{x + 7}{6x - 12} = \frac{2x + 1}{2(x - 2)} - \frac{2x - 1}{3(x - 2)} - \frac{x + 7}{6(x - 2)} = \frac{3(2x + 1) - 2(2x - 1) - (x + 7)}{6(x - 2)} = \frac{6x + 3 - 4x + 2 - x - 7}{6(x - 2)} = \frac{x - 2}{6(x - 2)} = \frac{1}{6}$
2) $\frac{24 - 2a}{a^2 - 16} - \frac{a}{2a - 8} + \frac{4}{a + 4} = \frac{24 - 2a}{(a - 4)(a + 4)} - \frac{a}{2(a - 4)} + \frac{4}{a + 4} = \frac{2(24 - 2a) - a(a + 4) + 4 * 2(a - 4)}{2(a - 4)(a + 4)} = \frac{48 - 4a - a^2 - 4a + 8a - 32}{2(a - 4)(a + 4)} = \frac{16 - a^2}{2(a - 4)(a + 4)} = -\frac{a^2 - 16}{2(a - 4)(a + 4)} = -\frac{(a^2 - 16)}{2(a^2 - 16)} = -\frac{1}{2}$
115. Представьте в виде дроби выражение:
1) $1 - a + \frac{a^2 - 2}{a + 2}$;
2) $\frac{a^2 - b^2}{3a + b} + 3a - b$;
3) $\frac{c^2 + 9}{c - 3} - c - 3$;
4) $\frac{8m^2}{4m - 3} - 2m - 1$.
Решение:
1) $1 - a + \frac{a^2 - 2}{a + 2} = \frac{a + 2 - a(a + 2) + a^2 - 2}{a + 2} = \frac{a + 2 - a^2 - 2a + a^2 - 2}{a + 2} = \frac{-a}{a + 2} = -\frac{a}{a + 2}$
2) $\frac{a^2 - b^2}{3a + b} + 3a - b = \frac{a^2 - b^2 + 3a(3a + b) - b(3a + b)}{3a + b} = \frac{a^2 - b^2 + 9a^2 + 3ab - 3ab - b^2}{3a + b} = \frac{10a^2 - 2b^2}{3a + b}$
3) $\frac{c^2 + 9}{c - 3} - c - 3 = \frac{c^2 + 9 - c(c - 3) - 3(c - 3)}{c - 3} = \frac{c^2 + 9 - c^2 + 3c - 3c + 9}{c - 3} = \frac{18}{c - 3}$
4) $\frac{8m^2}{4m - 3} - 2m - 1 = \frac{8m^2 - 2m(4m - 3) - (4m - 3)}{4m - 3} = \frac{8m^2 - 8m^2 + 6m - 4m + 3}{4m - 3} = \frac{2m + 3}{4m - 3}$
116. Упростите выражение:
1) $b + 7 - \frac{14b}{b + 7}$;
2) $5c - \frac{10 - 29c + 10c^2}{2c - 5} + 2$.
Решение:
1) $b + 7 - \frac{14b}{b + 7} = \frac{b(b + 7) + 7(b + 7) - 14b}{b + 7} = \frac{b^2 + 7b + 7b + 49 - 14b}{b + 7} = \frac{b^2 + 49}{b + 7}$
2) $5c - \frac{10 - 29c + 10c^2}{2c - 5} + 2 = \frac{5c(2c - 5) - (10 - 29c + 10c^2) + 2(2c - 5)}{2c - 5} = \frac{10c^2 - 25c - 10 + 29c - 10c^2 + 4c - 10}{2c - 5} = \frac{8c - 20}{2c - 5} = \frac{4(2c - 5)}{2c - 5} = 4$
117. Упростите выражение и найдите его значение:
1) $\frac{7}{2a - 4} - \frac{12}{a^2 - 4} - \frac{3}{a + 2}$, если a = 5;
2) $\frac{2c + 3}{2c^2 - 3c} + \frac{2c - 3}{2c^2 + 3c} - \frac{16c}{4c^2 - 9}$, если c = −0,8;
3) $\frac{m^2 + 16n^2}{m^2 - 16n^2} - \frac{m + 4n}{2m - 8n}$, если m = 3, n = 0,5.
Решение:
1) $\frac{7}{2a - 4} - \frac{12}{a^2 - 4} - \frac{3}{a + 2} = \frac{7}{2(a - 2)} - \frac{12}{(a - 2)(a + 2)} - \frac{3}{a + 2} = \frac{7(a + 2) - 2 * 12 - 2 * 3(a - 2)}{2(a - 2)(a + 2)} = \frac{7a + 14 - 24 - 6a + 12}{2(a - 2)(a + 2)} = \frac{a + 2}{2(a - 2)(a + 2)} = \frac{1}{2(a - 2)}$
при a = 5:
$\frac{1}{2(a - 2)} = \frac{1}{2(5 - 2)} = \frac{1}{2 * 3} = \frac{1}{6}$
2) $\frac{2c + 3}{2c^2 - 3c} + \frac{2c - 3}{2c^2 + 3c} - \frac{16c}{4c^2 - 9} = \frac{2c + 3}{с(2с - 3)} + \frac{2c - 3}{с(2с + 3)} - \frac{16c}{(2с - 3)(2с + 3)} = \frac{(2c + 3)(2c + 3) + (2c - 3)(2c - 3) - 16c * c}{c(2с - 3)(2с + 3)} = \frac{(2c + 3)^2 + (2c - 3)^2 - 16c^2}{c(2с - 3)(2с + 3)} = \frac{4c^2 + 12c + 9 + 4c^2 - 12c + 9 - 16c^2}{c(2с - 3)(2с + 3)} = \frac{-8c^2 + 18}{c(2с - 3)(2с + 3)} = \frac{-(8c^2 - 18)}{c(2с - 3)(2с + 3)} = \frac{-(8c^2 - 18)}{c(2с - 3)(2с + 3)} = \frac{-2(4c^2 - 9)}{c(4c^2 - 9)} = \frac{-2}{c}$
при c = −0,8:
$\frac{-2}{c} = \frac{-2}{-0,8} = \frac{2}{\frac{8}{10}} = 2 * \frac{5}{4} = \frac{5}{2} = 2,5$
3) $\frac{m^2 + 16n^2}{m^2 - 16n^2} - \frac{m + 4n}{2m - 8n} = \frac{m^2 + 16n^2}{(m - 4n)(m + 4n)} - \frac{m + 4n}{2(m - 4n)} = \frac{2(m^2 + 16n^2) - (m + 4n)(m + 4n)}{2(m - 4n)(m + 4n)} = \frac{2m^2 + 32n^2 - (m + 4n)^2}{2(m - 4n)(m + 4n)} = \frac{2m^2 + 32n^2 - (m^2 + 8n + 16n^2)}{2(m - 4n)(m + 4n)} = \frac{2m^2 + 32n^2 - m^2 - 8n - 16n^2}{2(m - 4n)(m + 4n)} = \frac{m^2 - 8n + 16n^2}{2(m - 4n)(m + 4n)} = \frac{(m - 4n)^2}{2(m - 4n)(m + 4n)} = \frac{m - 4n}{2(m + 4n)}$
при m = 3, n = 0,5:
$\frac{m - 4n}{2(m + 4n)} = \frac{3 - 4 * 0,5}{2(3 + 4 * 0,5)} = \frac{3 - 2}{2(3 + 2)} = \frac{1}{2 * 5} = \frac{1}{10} = 0,1$
118. Найдите значение выражения:
1) $\frac{6}{5x - 20} - \frac{x - 5}{x^2 - 8x + 16}$, если x = 5;
2) $\frac{2y - 1}{2y} - \frac{2y}{2y - 1} - \frac{1}{2y - 4y^2}$, если $y = -2\frac{3}{7}$.
Решение:
1) $\frac{6}{5x - 20} - \frac{x - 5}{x^2 - 8x + 16} = \frac{6}{5(x - 4)} - \frac{x - 5}{(x - 4)^2} = \frac{6(x - 4) - 5(x - 5)}{5(x - 4)^2} = \frac{6x - 24 - 5x + 25}{5(x - 4)^2} = \frac{x + 1}{5(x - 4)^2}$
при x = 5:
$\frac{x + 1}{5(x - 4)^2} = \frac{5 + 1}{5(5 - 4)^2} = \frac{6}{5 * 1^2} = \frac{6}{5} = 1\frac{1}{5}$
2) $\frac{2y - 1}{2y} - \frac{2y}{2y - 1} - \frac{1}{2y - 4y^2} = \frac{2y - 1}{2y} - \frac{2y}{2y - 1} + \frac{1}{4y^2 - 2y} = \frac{2y - 1}{2y} - \frac{2y}{2y - 1} + \frac{1}{2y(2y - 1)} = \frac{(2y - 1)^2 - 2y * 2y + 1}{2y(2y - 1)} = \frac{4y^2 - 4y + 1 - 4y^2 + 1}{2y(2y - 1)} = \frac{-4y + 2}{2y(2y - 1)} = \frac{-2(2y - 1)}{2y(2y - 1)} = -\frac{1}{y}$
при $y = -2\frac{3}{7}$:
$\frac{1}{y} = -\frac{1}{-2\frac{3}{7}} = -\frac{1}{-\frac{17}{7}} = \frac{7}{17}$
119. Докажите тождество:
1) $\frac{a + b}{a} - \frac{a}{a - b} + \frac{b^2}{a^2 - ab} = 0$;
2) $\frac{a + 3}{a + 1} - \frac{a + 1}{a - 1} + \frac{6}{a^2 - 1} = \frac{2}{a^2 - 1}$;
3) $\frac{2a^2 + 4}{a^2 - 1} - \frac{a - 2}{a + 1} - \frac{a + 1}{a - 1} = \frac{1}{a - 1}$.
Решение:
1) $\frac{a + b}{a} - \frac{a}{a - b} + \frac{b^2}{a^2 - ab} = 0$
$\frac{a + b}{a} - \frac{a}{a - b} + \frac{b^2}{a(a - b)} = 0$
$\frac{(a + b)(a - b) - a * a + b^2}{a(a - b)} = 0$
$\frac{a^2 - b^2 - a^2 + b^2}{a(a - b)} = 0$
$\frac{0}{a(a - b)} = 0$
0 = 0
2) $\frac{a + 3}{a + 1} - \frac{a + 1}{a - 1} + \frac{6}{a^2 - 1} = \frac{2}{a^2 - 1}$
$\frac{a + 3}{a + 1} - \frac{a + 1}{a - 1} + \frac{6}{(a - 1)(a + 1)} = \frac{2}{a^2 - 1}$
$\frac{(a + 3)(a - 1) - (a + 1)(a + 1) + 6}{(a - 1)(a + 1)} = \frac{2}{a^2 - 1}$
$\frac{a^2 + 3a - a - 3 - (a^2 + 2a + 1) + 6}{(a - 1)(a + 1)} = \frac{2}{a^2 - 1}$
$\frac{a^2 + 2a - 3 - a^2 - 2a - 1 + 6}{(a - 1)(a + 1)} = \frac{2}{a^2 - 1}$
$\frac{2}{a^2 - 1} = \frac{2}{a^2 - 1}$
3) $\frac{2a^2 + 4}{a^2 - 1} - \frac{a - 2}{a + 1} - \frac{a + 1}{a - 1} = \frac{1}{a - 1}$
$\frac{2a^2 + 4}{(a - 1)(a + 1)} - \frac{a - 2}{a + 1} - \frac{a + 1}{a - 1} = \frac{1}{a - 1}$
$\frac{2a^2 + 4 - (a - 2)(a - 1) - (a + 1)(a + 1)}{(a - 1)(a + 1)} = \frac{1}{a - 1}$
$\frac{2a^2 + 4 - (a^2 - 2a - a + 2) - (a^2 + a + a + 1)}{(a - 1)(a + 1)} = \frac{1}{a - 1}$
$\frac{2a^2 + 4 - a^2 + 2a + a - 2 - a^2 - a - a - 1}{(a - 1)(a + 1)} = \frac{1}{a - 1}$
$\frac{a + 1}{(a - 1)(a + 1)} = \frac{1}{a - 1}$
$\frac{1}{a - 1} = \frac{1}{a - 1}$
