1/ Cho biểu thức:
\(Q=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right).\left(\dfrac{x+x\sqrt{x}}{\sqrt{x}-1}-\dfrac{1-\sqrt{x}}{\sqrt{x}-x}\right)\)với x>0, x\(\ne\)1
a) rút gọn Q
b) Tìm các giá trị của x để Q= -1
2/ Thu gọn biểu thức sau:
a) \(A=\dfrac{5+\sqrt{5}}{\sqrt{5}+2}+\dfrac{\sqrt{5}}{\sqrt{5}-1}-\dfrac{3\sqrt{5}}{3+\sqrt{5}}\)
b) \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)( x >0)
Giúp mk với
Bài 2:
a: \(A=\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{4}-\dfrac{3\sqrt{5}\left(3-\sqrt{5}\right)}{4}\)
\(=-5+3\sqrt{5}+\dfrac{5+\sqrt{5}-9\sqrt{5}+15}{4}\)
\(=-5+3\sqrt{5}+5-2\sqrt{5}=\sqrt{5}\)
b: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+3\sqrt{x}+6-2\sqrt{x}-6}=1\)