ПОМОГИТЕ ПОЖАЛУЙСТА:С на фотографии номер 7.. Created by AngelinkaSo. algebra-ru

ПОМОГИТЕ ПОЖАЛУЙСТА:С на фотографии номер 7...

AngelinkaSo @AngelinkaSo

June 2022 1 32 Report

ПОМОГИТЕ ПОЖАЛУЙСТА:С
на фотографии номер 7.

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ndusha

Verified answer

$( \frac{x+1}{x-1}- \frac{x-1}{x+1}):( \frac{x^2+1}{x^2-1}- \frac{x^2-1}{x^2+1}) = \\ ( \frac{(x+1)^2}{(x-1)(x+1)}- \frac{(x-1)^2}{(x+1)(x-1)}):( \frac{(x^2+1)^2}{(x^2-1)(x^2+1)}- \frac{(x^2-1)^2}{(x^2-1)(x^2+1)}) = \\ \frac{(x+1)^2-(x-1)^2}{x^2-1}: \frac{(x^2+1)^2-(x^2-1)^2}{(x^2-1)(x^2+1)} = \\ \frac{(x+1)^2-(x-1)^2}{x^2-1}* \frac{(x^2-1)(x^2+1)}{(x^2+1)^2-(x^2-1)^2} = \\ \frac{(x+1)^2-(x-1)^2}{1}* \frac{(x^2+1)}{(x^2+1)^2-(x^2-1)^2} = \\$
$\frac{1}{1}* \frac{(x^2+1)}{1}=x^2+1$
при х=-3 3/4=-15/4
x²+1=(-15/4)²+1=225/16+1=14,0625+1=15,0625

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