прошу помочь с алгеброй. Created by melnik210. algebra-ru

прошу помочь с алгеброй...

July 2022 1 4 Report

прошу помочь с алгеброй

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математик2011 $( \frac{3a+1}{a-1} -a)* \frac{6}{a+1} = \frac{3a+1+a(a-1)}{a-1} * \frac{6}{a+1} = \frac{(3a+1+ a^{2}-a )*6}{(a-1)(a+1)} = \frac{6*( a^{2} +2a+1)}{(a-1)(a+1)} = \\ \\ = \frac{6 (a+1)^{2} }{(a-1)(a+1)} = \frac{6(a+1)}{a-1}$

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