Question

Помогите пожалуйста, задание на фото

Answer 1

Ответ:

$1)\ \ \underline{\ Esli\ a<b\ ,\ to\ \sqrt{a}<\sqrt{b}\ }.\\\\81<87<100\ \ \ \Rightaeeow\ \ \ \sqrt{81}<\sqrt{87}<\sqrt{100}\ \ ,\ \ \underline{9<\sqrt{87}<10}\\\\\\16<17<25\ \ \ \Rightaeeow\ \ \ \sqrt{16}<\sqrt{17}<\sqrt{25}\ \ ,\ \ 4<\sqrt{17}<5\ \ \to \\\\4+2<\sqrt{17}+2<5+2\ \ ,\ \ \ \underline{6<\sqrt{17}+2<7\ }$

$2)\ \ \underline{\sqrt{a^2}=|a|}=\left\{\begin{array}{l}a,\ esli\ a\geq 0\ ,\\-a,\ esli\ a<0\ .\end{array}\right\\\\\sqrt{8x}+\sqrt{32x}-\sqrt{98x}=\sqrt{2^2\cdot 2x}+\sqrt{4^2\cdot 2x}-\sqrt{7^2\cdot 2x}=\\\\=2\sqrt{2x}+4\sqrt{2x}-7\sqrt{2x}=(2+4-7)\cdot \sqrt{2x}=-\sqrt{2x}$

$3)\ \ \underline {a<b\ \ \Rightarrow \ \ \ -b<-a}\\\\-6\sqrt2=-\sqrt{6^2\cdot 2}=-\sqrt{72}\\\\-4\sqrt7=-\sqrt{4^2\cdot 7} =-\sqrt{112}\\\\-5\sqrt3=-\sqrt{5^2\cdot 3}=-\sqrt{75}\\\\72<75<112\ \ \Rightarrow \ \ \ \sqrt{72}<\sqrt{75}<\sqrt{112}\ \ \ \Rightarrow \ \ \ -\sqrt{112}<-\sqrt{75}<-\sqrt{72}\ \to \\\\\underline {-4\sqrt7<-5\sqrt3<-6\sqrt2\ }$