Решебник по алгебре 9 класс Макарычев Задание 922

 

\[\boxed{\text{922\ (922).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \frac{5 + \sqrt{y}}{5\sqrt{y} + y} = \frac{5 + \sqrt{y}}{\sqrt{y} \cdot \left( 5 + \sqrt{y} \right)} =\]

\[= \frac{1}{\sqrt{y}} = \frac{\sqrt{y}}{y}\]

\[\textbf{б)}\ \frac{3x - 6}{\sqrt{x} + \sqrt{2}} =\]

\[= \frac{3 \cdot \left( \sqrt{x} - \sqrt{2} \right)\left( \sqrt{x} + \sqrt{2} \right)}{\left( \sqrt{x} + \sqrt{2} \right)} =\]

\[= 3\sqrt{x} - 3\sqrt{2}\]

\[\textbf{в)}\ \frac{a\sqrt{a} - 1}{a + \sqrt{a} + 1} = \frac{\sqrt{a^{3}} - 1}{a + \sqrt{a} + 1} =\]

\[= \frac{\left( \sqrt{a} \right)^{3} - 1^{3}}{a + \sqrt{a} + 1} =\]

\[= \frac{\left( \sqrt{a} - 1 \right)\left( a + \sqrt{a} + 1 \right)}{a + \sqrt{a} + 1} =\]

\[= \sqrt{a} - 1\]

\[\textbf{г)}\ \frac{b - \sqrt{b} + 1}{b\sqrt{b} + 1} = \frac{b - \sqrt{b} + 1}{\sqrt{b^{3}} + 1} =\]

\[= \frac{b - \sqrt{b} + 1}{\left( \sqrt{b} \right)^{3} + 1^{3}} =\]

\[= \frac{b - \sqrt{b} + 1}{\left( \sqrt{b} + 1 \right)\left( b - \sqrt{b} + 1 \right)} =\]

\[= \frac{1}{\sqrt{b} + 1}\]

\[\textbf{д)}\ \frac{x\sqrt{x} + y\sqrt{y}}{\sqrt{\text{xy}} + y} =\]

\[= \frac{\sqrt{x^{3}} + \sqrt{y^{3}}}{\sqrt{y} \cdot \left( \sqrt{x} + \sqrt{y} \right)} =\]

\[= \frac{\left( \sqrt{x} \right)^{3} + \left( \sqrt{y} \right)^{3}}{\sqrt{y} \cdot \left( \sqrt{x} + \sqrt{y} \right)} =\]

\[= \frac{\left( \sqrt{x} + \sqrt{y} \right)\left( x - \sqrt{\text{xy}} + y \right)}{\sqrt{y} \cdot \left( \sqrt{x} + \sqrt{y} \right)} =\]

\[= \frac{x - \sqrt{\text{xy}} + y}{\sqrt{y}}\]

\[\textbf{е)}\ \frac{c - \sqrt{\text{cd}}}{c\sqrt{c} - d\sqrt{d}} =\]

\[= \frac{\sqrt{c} \cdot \left( \sqrt{c} - \sqrt{d} \right)}{\sqrt{c^{3}} - \sqrt{d^{3}}} =\]

\[= \frac{\sqrt{c} \cdot \left( \sqrt{c} - \sqrt{d} \right)}{\left( \sqrt{c} \right)^{3} - \left( \sqrt{d} \right)^{3}} =\]

\[= \frac{\sqrt{c} \cdot \left( \sqrt{c} - \sqrt{d} \right)}{\left( \sqrt{c} - \sqrt{d} \right)\left( c + \sqrt{\text{cd}} + d \right)} =\]

\[= \frac{\sqrt{c}}{c + \sqrt{\text{cd}} + d}.\]