Question

Помогите пожалуйста даю 20 баллов

Answer 1

$1)\ \ \dfrac{x-4}{\sqrt{x}-2}=\dfrac{\sqrt{x}-2)(\sqrt{x}+2)}{\sqrt{x}-2}=\sqrt{x}+2\\\\\\2)\ \ \dfrac{x+\sqrt{x}}{x-1}=\dfrac{\sqrt{x}\, (\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\\\\\\3)\ \ \dfrac{4\sqrt{x}-x}{2x-32}=\dfrac{\sqrt{x}\, (4-\sqrt{x})}{2(x-16)}=\dfrac{\sqrt{x}\, (4-\sqrt{x})}{2(\sqrt{x}-4)(\sqrt{x}+4)}=\dfrac{-\sqrt{x}}{2(\sqrt{x}+4)}$

$4)\ \ \dfrac{\sqrt5+5}{4\sqrt5}=\dfrac{\sqrt5\, (1+\sqrt5)}{4\sqrt5}=\dfrac{1+\sqrt5}{4}\\\\\\5)\ \ \dfrac{9b-3}{3\sqrt{b}+\sqrt3}=\dfrac{3(\sqrt{3b}-1)(\sqrt{3b}+1)}{\sqrt3\, (\sqrt{3b}+1)}=\sqrt{3b}-1$

$6)\ \ \dfrac{\sqrt3-3}{3\sqrt2-\sqrt6}=\dfrac{\sqrt3\, (1-\sqrt3)}{\sqrt6\, (\sqrt3-1)}=\dfrac{-\sqrt3}{\sqrt6}=-\dfrac{1}{\sqrt2}\\\\\\7)\ \ \dfrac{9a-b^2}{9a-6b\sqrt{a}+b^2}=\dfrac{(3\sqrt{a}-b)(3\sqrt{a}+b)}{(3\sqrt{a}-b)^2}=\dfrac{3\sqrt{a}+b}{3\sqrt{a}-b}$

$8)\ \ \dfrac{5-\sqrt{15}}{\sqrt{15}-3}=\dfrac{\sqrt5\, (\sqrt5-\sqrt3)}{\sqrt3\, (\sqrt5-\sqrt3)}=\sqrt{\dfrac{5}{3}}\\\\\\9)\ \ \dfrac{a+2\sqrt{a}+4}{a\sqrt{a}-8}=\dfrac{a+2\sqrt{a}+4}{(\sqrt{a}-2)(a+2\sqrt{a}+4)}=\dfrac{1}{\sqrt{a}-2}$