Question

Срочно алгебра 8 класс​

Answer 1

Ответ:              $\boxed{\ \sqrt{x^2}=|x|\geq 0\ }$

$\boxed{\ \sqrt{x}\geq 0\ }\ \ \ \Rightarrow \ \ \ \sqrt{81}=9\ne -9\\\\\\\boxed{\ \sqrt{x^2}=|x|\ } \ \Rightarrow \\\\-\sqrt{0,01}=-1\cdot \sqrt{(0,1)^2}=-1\cdot |0,1|=-1\cdot 0,1=-0,1\ne 0,1\\\\\\\boxed{\ \sqrt{x^2}=|x|\ }\ \ \ \Rightarrow \ \ \ \sqrt{(-12)^2}=|-12|=12\ne -12\\\\\\\boxed{\ \sqrt{x}\ ,\ x\geq 0\ }\ \ \Rightarrow \ \sqrt{-2^{-2}}=\sqrt{-1\cdot 2^{-2}}=\sqrt{-(2^{-2})}=\sqrt{-\dfrac{1}{4}}\ \ \ ne\ \ syshestvyet\ ,\\\\tak\ kak\ \ -\dfrac{1}{4}<0$