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МатематикаРешебник по алгебре 8 класс Макарычев ФГОС Задание 256
Задание 256
\[\boxed{\text{256\ (256).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ y = \frac{36}{(x + 1)^{2} - (x - 1)^{2}} =\]
\[= \frac{36}{x^{2} + 2x + 1 - x^{2} + 2x - 1} =\]
\[= \frac{36}{4x} = \frac{9}{x}\]
\[ООФ:\]
\[(x + 1)^{2} - (x - 1)^{2} \neq 0\]
\[(x + 1)^{2} \neq (x - 1)^{2}\]
\[x^{2} + 2x + 1 \neq x^{2} - 2x + 1\]
\[4x \neq 0\]
\[x \neq 0.\]
\[y = \frac{9}{x}.\]
\[\textbf{б)}\ y = \frac{18 - 12x}{x^{2} - 3x} - \frac{6}{3 - x} =\]
\[= \frac{18 - 2x}{x(x - 3)} + \frac{6^{\backslash x}}{x - 3} =\]
\[= \frac{18 - 12x + 6x}{x(x - 3)} =\]
\[= \frac{18 - 6x}{x(x - 3)} = \frac{6(3 - x)}{- x(3 - x)} = - \frac{6}{x}\]
\[ООФ:\]
\[x^{2} - 3x \neq 0\]
\[x(x - 3) \neq 0\]
\[x \neq 0;\ \ \ x \neq 3.\]
\[y = - \frac{6}{x}\]
\[\textbf{в)}\ y = \frac{16}{(2 - x)^{2} - (2 + x)^{2}} =\]
\[= \frac{16}{(2 - x - 2 - x)(2 - x + 2 + x)} =\]
\[= \frac{16}{- 2x \cdot 4} = - \frac{2}{x}\]
\[ООФ:\]
\[(2 - x)^{2} - (2 + x)^{2} \neq 0\]
\[(2 - x)^{2} \neq (2 + x)^{2}\]
\[4 - 2x + x^{2} \neq 4 + 2x + x^{2}\]
\[4x \neq 0\]
\[x \neq 0.\]
\[y = - \frac{2}{x}\]
\[\textbf{г)}\ \ y = \frac{3x(x + 1) - 3x^{2} + 15}{x(x + 5)} =\]
\[= \frac{3x^{2} + 3x - 3x^{2} + 15}{x(x + 5)} =\]
\[= \frac{3(x + 5)}{x(x + 5)} = \frac{3}{x}\]
\[ООФ:\ \ \]
\[x(x + 5) \neq 0\]
\[x \neq 0;\ \ x \neq - 5.\ \ \]
\[y = \frac{3}{x};\ \ x \neq - 5\]
\[\boxed{\text{256.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]
\[\frac{1}{z - a} + \frac{1}{z - b} = \frac{1}{a} + \frac{1}{b}\]
\[Доказать,\ что\ z = \frac{2}{\frac{1}{a} + \frac{1}{b}}.\]
\[\frac{1}{\frac{2}{\frac{1^{\backslash a}}{a} + \frac{1^{\backslash b}}{b}} - a} + \frac{1}{\frac{2}{\frac{1^{\backslash b}}{a} + \frac{1^{\backslash a}}{b}} - b} =\]
\[= \frac{1}{a} + \frac{1}{b}\]
\[\frac{1}{\frac{2ab}{b + a} - a^{\backslash b + a}} + \frac{1}{\frac{2ab}{b + a} - b^{\backslash b + a}} =\]
\[= \frac{1}{a} + \frac{1}{b}\]
\[\frac{b + a}{2ab - ab - a^{2}} + \frac{b + a}{2ab - ab - b^{2}} =\]
\[= \frac{1}{a} + \frac{1}{b}\]
\[\frac{b + a^{\backslash b}}{ab - a^{2}} + \frac{b + a^{\backslash a}}{ab - b^{2}} = \frac{1}{a} + \frac{1}{b}\]
\[\frac{b + a}{\text{ab}} = \frac{1}{a} + \frac{1}{b}\]
\[\frac{a}{\text{ab}} + \frac{b}{\text{ab}} = \frac{1}{a} + \frac{1}{b}\]
\[\frac{1}{a} + \frac{1}{b} = \frac{1}{a} + \frac{1}{b}\]
\[Что\ и\ требовалось\ доказать.\]
