99 БАЛЛОВ!!!!!!!!! Упростите выражение [tex] \frac{c+5}{c^2-64} : ( \frac{4}{c+8} - \frac{12}{c^2+16c+64} )+ \frac{4}{8-c} [/tex]. Created by ХомячокРазумный. algebra-ru

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August 2022 1 6 Report

99 БАЛЛОВ!!!!!!!!! Упростите выражение
[tex] \frac{c+5}{c^2-64} : ( \frac{4}{c+8} - \frac{12}{c^2+16c+64} )+ \frac{4}{8-c} [/tex]

Answers & Comments

wangross $\frac{c+5}{c^2-64} : ( \frac{4}{c+8} - \frac{12}{c^2+16c+64} )+ \frac{4}{8-c}= \\ \\ = \frac{c+5}{c^2-64}:( \frac{4}{c+8}- \frac{12}{(c+8)^2})+ \frac{4}{8-c}= \\ \\=\frac{c+5}{c^2-64}: \frac{4(c+8)-12}{(c+8)^2}+ \frac{4}{8-c}= \\ \\=\frac{c+5}{c^2-64}: \frac{4c+32-12}{(c+8)^2} + \frac{4}{8-c}= \\ \\=\frac{c+5}{c^2-64}:\frac{4c+20}{(c+8)^2}+ \frac{4}{8-c} = \frac{(c+5)*(c+8)^2}{(c-8)(c+8)*4(c+5)}- \frac{4}{c-8}= \\ \\ = \frac{c+8}{4(c-8)} - \frac{4}{c-8}= \frac{c+8-16}{4(c-8)}=$

$=\frac{c-8}{4(c-8)} = \frac{1}{4} =0,25$

Ответ: $0,25$

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