Question

Помогите решить срочно!!!! ​

Answer 1

Пошаговое объяснение:

$\tt\displaystyle \frac{{{x^2}+14x+49}}{{x+6}}:\left({\frac{{13}}{{x+6}}-x+6} \right)=\frac{{{{(x+7)}^2}}}{{x+6}}:\left( {\frac{{13}}{{x+6}}-\frac{{x(x+6)}}{{x+6}}+\frac{{6(x+6)}}{{x+6}}} \right)=$

$\[\begin{gathered}=\frac{{{{(x+7)}^2}}}{{x+6}}:\left( {\frac{{13-{x^2}-6x+6x+36}}{{x+6}}} \right)=\frac{{{{(x+7)}^2}}}{{x+6}}:\left({\frac{{-{x^2}+49}}{{x+6}}}\right)=\hfill \\\\=\frac{{{{(x+7)}^2}}}{{x+6}}*\frac{{x+6}}{{(49-{x^2})}} =\frac{{{{(x+7)}^2}}}{{(7-x)(7+x)}}=\boxed{\frac{{(x+7)}}{{(7-x)}}} \hfill \\ \end{gathered} \]$

$\tt\displaystyle \left( {c-\frac{{2c-9}}{{c+8}}}\right):\frac{{{c^2}+3c}}{{{c^2}- 64}}+\frac{{24}}{c}=\left({\frac{{c(c+8)}}{{c+8}}-\frac{{2c-9}}{{c+8}}}\right):\frac{{c(c+3)}}{{(c-8)(c+8)}}+\frac{{24}}{c}=\left( {\frac{{{c^2}+8c-2c+9}}{{c+8}}}\right)*\frac{{(c-8)(c+8)}}{{c(c+3)}}+\frac{{24}}{c}=\frac{{{c^2}+6c+9}}{{c+8}}*\frac{{(c-8)(c+8)}}{{c(c+3)}}+\frac{{24}}{c}=\frac{{{{(c+3)}^2}}}{{c+8}}*\frac{{(c-8)(c+8)}}{{c(c+3)}}+\frac{{24}}{c}=\frac{{(c+3)(c-8)}}{c}+\frac{{24}}{c}=\frac{{24+(c+3)(c-8)}}{c}=\frac{{24+{c^2}-5c-24}}{c}=\frac{{{c^2}-5c}}{c}=\frac{{c(c-5)}}{c}=\boxed{c-5}$