Question

помогите пожалуйста! Даю 35 баллов!

Answer 1

Ответ:

Объяснение:

$\frac{c^2+6c+9}{c} :\big (1+\frac{3}{c} \big)=\frac{(c+3)^2}{c} :\big (\frac{c+3}{c} \big)=\frac{(c+3)^2}{c} * \frac{c}{c+3} =c+3=-11+3=\fbox{-8}$

$\sqrt{x} \big(\sqrt{x} -\frac{1}{y} \big)+1:\frac{y}{\sqrt{x} } =x-\frac{\sqrt{x} }{y} +1*\frac{\sqrt{x} }{y}=x=\fbox{123}$

Answer 2

$1)\dfrac{c^{2}+6c+9 }{c}:\Big(1+\dfrac{3}{c}\Big)=\dfrac{(c+3)^{2}}{c} :\dfrac{c+3}{c}=\dfrac{(c+3)^{2}}{c}\cdot \dfrac{c}{c+3}=c+3\\\\c=-11 \ \Rightarrow \ c+3=-11+3=\boxed{-8}\\\\\\2)\sqrt{x} \Big(\sqrt{x}-\dfrac{1}{y}\Big)+1:\dfrac{y}{\sqrt{x} } =x-\dfrac{\sqrt{x} }{y}+1\cdot \dfrac{\sqrt{x} }{y} =x-\dfrac{\sqrt{x} }{y}+\dfrac{\sqrt{x} }{y}=x\\\\\boxed{x=123}$