Question

СРОЧНО ПОМОГИТЕ!!!
алгебра ​

Answer 1

Ответ:

$1)\ \ \underline{\sqrt{a^2}=|a|\ }\\\\\sqrt{72}=\sqrt{36\cdot 2}=\sqrt{6^2\cdot 2}=6\sqrt2\\\\\sqrt{50}=\sqrt{5^2\cdot 2}=5\sqrt2\\\\\sqrt{125}=\sqrt{5^2\cdot 5}=5\sqrt5\\\\\sqrt{567}=\sqrt{9^2\cdot 7}=9\sqrt7\\\\\\2)\ \ \underline{\ a\sqrt{b}=\sqrt{a^2b}\ ,\ esli\ \ a\geq 0\ }\\\\7\sqrt5=\sqrt{7^2\cdot 5}=\sqrt{245}\\\\-9\sqrt7=-\sqrt{9^2\cdot 7}=-\sqrt{567}\\\\\dfrac{2}{3}\sqrt{27}=\sqrt{\dfrac{4}{9}\cdot 27}=\sqrt{4\cdot 3}=\sqrt{12}$

$3)\ \ \underline {a<b\ \ \ \Rightarrow \ \ \ \sqrt{a}<\sqrt{b}\ }\\\\6\sqrt2=\sqrt{6^2\cdot 2}=\sqrt{72}\ \ ,\\\\72<82\ \ \ \Rightarrow \ \ \ \ \ \sqrt{72}<\sqrt{82}\ \ \ \Rightarrow \ \ \ \ 6\sqrt2<\sqrt{82}\\\\\\b)\ \ \dfrac{3}{5}\sqrt{75}=\sqrt{\dfrac{9\cdot 75}{25}}=\sqrt{9\cdot 3}=\sqrt{27}\\\\10\sqrt{\dfrac{3}{5}}=\sqrt{\dfrac{3\cdot 100}{5}}=\sqrt{60}\\\\27<60\ \ \Rightarrow \ \ \ \sqrt{27}<\sqrt{60}\ \ \ \Rightarrow \ \ \ \ \dfrac{3}{5}\sqrt{75}<10\sqrt{\dfrac{3}{5}}$