СРОЧНО! 58, ПОМОГИТЕ, ПОЖАЛУЙСТА. Created by raikers. algebra-ru

СРОЧНО! 58, ПОМОГИТЕ, ПОЖАЛУЙСТА...

raikers @raikers

July 2022 1 39 Report

СРОЧНО! 58, ПОМОГИТЕ, ПОЖАЛУЙСТА

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ndusha

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$( \frac{2}{5 \sqrt{a}-25 } -\frac{4}{a-25 } -\frac{1}{5 \sqrt{a}+25 })*(a+10 \sqrt{a} +25)= \\ ( \frac{2}{5( \sqrt{a}-5) } -\frac{4}{ \sqrt{a}^2-5^2 } -\frac{1}{5( \sqrt{a}+5) })*( \sqrt{a}^2+2*5 \sqrt{a} +5^2)= \\ ( \frac{2( \sqrt{a}+5)}{5( \sqrt{a}-5)( \sqrt{a}+5) } -\frac{4*5}{5 ( \sqrt{a}-5)( \sqrt{a}+5)} -\frac{1( \sqrt{a}-5)}{5( \sqrt{a}+5)( \sqrt{a}-5) })*(\sqrt{a}+5)^2= \\$
$( \frac{2( \sqrt{a}+5)}{5( \sqrt{a}-5)( \sqrt{a}+5) } -\frac{4*5}{5 ( \sqrt{a}-5)( \sqrt{a}+5)} -\frac{1( \sqrt{a}-5)}{5( \sqrt{a}+5)( \sqrt{a}-5) })*( \sqrt{a}+5)^2 \\ \frac{2( \sqrt{a}+5)-20-( \sqrt{a}-5)}{5( \sqrt{a}+5)( \sqrt{a}-5) }*( \sqrt{a}+5)^2= \frac{2 \sqrt{a}+10-20- \sqrt{a}+5}{5( \sqrt{a}+5)( \sqrt{a}-5) }*( \sqrt{a}+5)^2= \\ \frac{ \sqrt{a}-5}{5( \sqrt{a}+5)( \sqrt{a}-5) }*( \sqrt{a}+5)^2= \frac{1}{5( \sqrt{a}+5)}*( \sqrt{a}+5)^2= \\ \frac{1}{5}*( \sqrt{a}+5)= \sqrt{a}/5+1= 0.2\sqrt{a}+1$

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