Решебник по алгебре 8 класс Мерзляк ФГОС Задание 1123

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\[1)\ \frac{x^{2} - 19}{x + \sqrt{19}} =\]

\[= \frac{\left( x - \sqrt{19} \right)\left( x + \sqrt{19} \right)}{\left( x + \sqrt{19} \right)} =\]

\[= x - \sqrt{19}\]

\[2)\ \frac{\sqrt{x} - 6}{x - 36} =\]

\[= \frac{(\sqrt{x} - 6)}{(\sqrt{x} - 6)(\sqrt{x} + 6)} = \frac{1}{\sqrt{x} + 6}\]

\[3)\ \frac{m + 8\sqrt{m}}{m - 64} =\]

\[= \frac{\sqrt{m} \cdot (\sqrt{m} + 8)}{(\sqrt{m} + 8)(\sqrt{m} - 8)} = \frac{\sqrt{m}}{\sqrt{m} - 8}\]

\[4)\ \frac{29 - \sqrt{29}}{\sqrt{29}} =\]

\[= \frac{\sqrt{29} \cdot (\sqrt{29} - 1)}{\sqrt{29}} = \sqrt{29} - 1\]

\[5)\ \frac{a - 6\sqrt{\text{ab}} + 9b}{a - 9b} =\]

\[= \frac{\left( \sqrt{a} - 3\sqrt{b} \right)^{2}}{\left( \sqrt{a} - 3\sqrt{b} \right)\left( \sqrt{a} + 3\sqrt{b} \right)} =\]

\[= \frac{\sqrt{a} - 3\sqrt{b}}{\sqrt{a} + 3\sqrt{b}}\]

\[6)\ \frac{11 - \sqrt{33}}{\sqrt{33} - 3} =\]

\[= \frac{\sqrt{11} \cdot (\sqrt{11} - \sqrt{3})}{\sqrt{3} \cdot (\sqrt{11} - \sqrt{3})} = \sqrt{\frac{11}{3}}\]