Question

№96,97,99 ПЖЖЖЖ ПОМОГИТЕ

Answer 1

Ответ:

$96)\ \ 5\sqrt{2}=\sqrt{50}\ \ ,\ \ 7\sqrt{3}=\sqrt{147}\ \ ,\ \ \ 0,3\sqrt6=\sqrt{0,54}\ \ ,\ \ \dfrac{1}{4}\sqrt{48}=\sqrt{3}\\\\\dfrac{2}{9}\sqrt{162}=\sqrt{8}\ \ ,\ \ -2\sqrt{10}=-\sqrt{40}\ \ ,\ \ -0,2\sqrt{30}=-\sqrt{12}\ \ ,\ \ 8\sqrt{b}=\sqrt{64b}$

$97)\ \ c\sqrt{15}=\sqrt{15c^2}\ \ ,\ \ c\geq 0\\\\m\geq 0\ ,\ \ m\sqrt{n}=\sqrt{m^2n}\\\\x^7\sqrt{-x}=-\sqrt{-x^{15}}\ \ ,\ \ x\leq 0\\\\8c\sqrt{\dfrac{c}{32}}=\sqrt{2c^3}\ \ ,\ \ c\geq 0\\\\(p+1)\sqrt{\dfrac{1}{p+1}}=\sqrt{\dfrac{(p+1)^2}{p+1}}=\sqrt{p+1}\ ,\ \ p>-1\\\\(b-4)\sqrt{\dfrac{1}{20-5b}}=-\sqrt{\dfrac{(b-4)^2}{-5(b-4)}}=-\sqrt{\dfrac{4-b}{5}}\ ,\ \ b<4$

$99)\ \ (\sqrt{45}+\sqrt{180})\cdot \sqrt{5}=\sqrt{45\cdot 5}+\sqrt{180\cdot 5}=15+30=45\\\\(6\sqrt2-6\sqrt{50}+\sqrt{72})\cdot \sqrt2=6\cdot 2-6\cdot 5\cdot 2+6\cdot 2=-36\\\\(4-\sqrt6)(2+3\sqrt6)=8+12\sqrt6-2\sqrt6-3\cdot 6=-10+10\sqrt6=10(\sqrt6-1)\\\\(2\sqrt7-\sqrt2)(\sqrt7+6\sqrt2)=14+12\sqrt2-\sqrt{14}-12=2+12\sqrt2-\sqrt{14}\\\\(\sqrt{19}-\sqrt{13})(\sqrt{19}+\sqrt{13})=19-13=6\\\\(6\sqrt{m}-8\sqrt{n})(6\sqrt{m}+8\sqrt{n})=36m-64n\\\\(\sqrt3+2)^2=7+4\sqrt3$

$\(2\sqrt6-3\sqrt7)^2=24-12\sqrt{42}+63=87-12\sqrt{42}$