Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 12

\[\boxed{\mathbf{12}.}\]

\[1)\ \frac{3x + 5}{x - 2} - \frac{11 - x}{x - 2} =\]

\[= \frac{3x + 5 - 11 + x}{x - 2} = \frac{4x - 6}{x - 2}\]

\[2)\ \frac{2a}{a - b} + \frac{2a - b}{b - a} = \frac{2a}{a - b} -\]

\[- \frac{2a - b}{a - b} = \frac{2a - 2a + b}{a - b} = \frac{b}{a - b}\]

\[3)\ \frac{3^{\backslash a - b}}{a} + 5^{\backslash a(a - b)} - \frac{2^{\backslash a}}{a - b} =\]

\[= \frac{3a - 3b + 5a^{2} - 5ab - 2a}{a(a - b)} =\]

\[= \frac{5a^{2} - 5ab - 3b + a}{a(a - b)}\]

\[4)\ \frac{3a}{6a + 8} - \frac{1}{2} - \frac{2}{4 - 3a} =\]

\[= \frac{3a^{\backslash 4 - 3a}}{2 \cdot (4 + 3a)} - \frac{1^{\backslash 16 - 9a^{2}}}{2} -\]

\[- \frac{2^{\backslash 2 \cdot (4 + 3a)}}{4 - 3a} =\]

\[= \frac{12a - 9a^{2} - 16 + 9a^{2} - 16 - 12a}{2 \cdot (4 + 3a)(4 - 3a)} =\]

\[= \frac{- 32}{2 \cdot \left( 16 - 9a^{2} \right)} = \frac{16}{9a^{2} - 16}\]

\[5)\ \frac{5b - b^{2}}{3a} \cdot \frac{6a^{2}}{b^{3} - 5b^{2}} =\]

\[= \frac{b(5 - b) \cdot 2a}{b^{2}(b - 5)} = - \frac{2a}{b}\]

\[6)\ \frac{6c^{3}}{9 - a^{2}} \cdot \frac{a^{2} - 6a + 9}{4a^{2}c} =\]

\[= \frac{3c^{2} \cdot (a - 3)^{2}}{- (a - 3)(a + 3) \cdot 2a^{2}} =\]

\[= - \frac{3c^{2}(a - 3)}{2a^{2}(a + 3)} =\]

\[= \frac{3c^{2}(3 - a)}{2a^{2}(3 + a)}\]

\[7)\ \frac{a^{3}b}{3a - 6b}\ :\frac{a^{2}b^{2} - a^{2}b}{ac - 2bc} =\]

\[= \frac{a^{3}b \cdot c \cdot (a - 2b)}{3 \cdot (a - 2b) \cdot a^{2}b(b - 1)} =\]

\[= \frac{\text{ac}}{3b - 3}\]

\[8)\ \frac{10 - 15b}{(a - b)^{2}}\ :\frac{9b^{2} - 4}{3b - 3a} =\]

\[= \frac{5 \cdot (2 - 3b) \cdot 3 \cdot (b - a)}{(a - b)^{2} \cdot (3b - 2)(3b + 2)} =\]

\[= \frac{15}{(a - b)(3b + 2)}\]

\[9)\ \left( \frac{a^{\backslash a + 3}}{7a - 4} - \frac{1^{\backslash 7a - 4}}{a + 3} \right) \cdot\]

\[\cdot \frac{12 - 21a}{(2 - a)^{2}} = \frac{a^{2} + 3a - 7a + 4}{(7a - 4)(a + 3)} \cdot\]

\[\cdot \frac{- 3 \cdot (7a - 4)}{(2 - a)^{2}} =\]

\[= \frac{a^{2} - 4a + 4}{a + 3} \cdot \frac{- 3}{(a - 2)^{2}} =\]

\[= \frac{- 3 \cdot (a - 2)^{2}}{(a + 3)(a - 2)^{2}} = - \frac{3}{a + 3}\]

\[10)\ \left( \frac{x}{x - 2} - \frac{x}{x + 2} - \frac{x^{2} + 4}{4 - x^{2}} \right)\ :\]

\[:\frac{2x + x^{2}}{(2 - x)^{2}} =\]

\[= \left( \frac{x^{\backslash x + 2}}{x - 2} - \frac{x^{\backslash x - 2}}{x + 2} + \frac{x^{2} + 4}{x^{2} - 4} \right) \cdot\]

\[\cdot \frac{(2 - x)^{2}}{2x + x^{2}} =\]

\[= \frac{x^{2} + 2x - x^{2} + 2x + x^{2} + 4}{(x - 2)(x + 2)} \cdot\]

\[\cdot \frac{(x - 2)^{2}}{x(2 + x)} =\]

\[= \frac{\left( x^{2} + 4x + 4 \right) \cdot (x - 2)}{x(x + 2)(x + 2)} =\]

\[= \frac{(x + 2)^{2}(x - 2)}{x(x + 2)^{2}} = \frac{x - 2}{x}\]