Rút Gọn
a) A = \(\dfrac{1}{\sqrt{x}+\sqrt{x-1}}\) \(-\)\(\dfrac{1}{\sqrt{x}-\sqrt{x-1}}\)\(-\)\(\dfrac{x\sqrt{x}-x}{1-\sqrt{x}}\)( Rút gọn và Tìm x để A > 0 )
b) B \(=\)\(\left(2-\dfrac{a-3\sqrt{a}}{\sqrt{a}-3}\right)\)\(.\)\(\left(2-\dfrac{5\sqrt{a}-\sqrt{ab}}{\sqrt{b}-5}\right)\)
c) C = \(\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}+2\right)\)\(.\)\(\left(2-\dfrac{\sqrt{x}+x}{1+\sqrt{x}}\right)\)
GIÚP MÌNH VS MAI MÌNH NỘP BÀI RỒI
b: \(B=\left(2-\dfrac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}-3}\right)\cdot\left(2-\dfrac{\sqrt{a}\left(5-\sqrt{b}\right)}{-\left(5-\sqrt{b}\right)}\right)\)
\(=\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)=4-a\)
c: \(C=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+2\right)\left(2-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\)
\(=\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)\)
=4-x
