tex]а=корень из 448 в=корень из 448...

August 2022 2 43 Report

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[tex]\frac{a^2-64b^2}{a^2} -\frac{a}{a-8b}[/tex]

а=корень из 448 в=корень из 448

ruslank1460

$\begin{matrix} \begin{matrix}a=\sqrt{448}\\b=\sqrt{448}\end{matrix} &\Rightarrow &a=b;a^2=b^2\end{matrix}\\\\\frac{a^2-64b^2}{a^2} -\frac{a}{a-8b} =\frac{a^2-64a^2}{a^2} -\frac{a}{a-8a} =\\1-64-\frac{1}{1-8} =-63+\frac{1}{7} =-62\frac{6}{7}\\\\Otvet:-62\frac{6}{7}$

Мы знаем чему равно а и условие неравенства знаменателя нулю можно не писать.

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$\frac{a^2-64b^2}{a^2} -\frac{a}{a-8b}=\frac{448-64\cdot 448}{448} -\frac{\sqrt{448}}{\sqrt{448}-8\sqrt{448}}=\frac{-63\cdot 448}{448} -\frac{\sqrt{448}}{-7\sqrt{448}}=-63+\frac{1}{7}=-62\frac{6}{7}$

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