mọi người ơi giúp mình với , mình đang cần rất gấp
Rút gọn bt sau
a)\(\dfrac{2x+1}{x\sqrt{x}-1}+\dfrac{1}{1-\sqrt{x}}\)
b)\(\dfrac{\sqrt{x}-4}{x-2\sqrt{x}}-\dfrac{3}{2-\sqrt{x}}\)
c)\(\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x-1}{\sqrt[]{x}-1}\)
d)\(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{3}{2-\sqrt{x}}+\dfrac{3\sqrt{x}-2}{x-4}\)
e)\(\dfrac{x+2\sqrt{x}-10}{x-\sqrt{x}-6}-\dfrac{\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}-2}\)
a: \(=\dfrac{2x+1-x-\sqrt{x}-1}{x\sqrt{x}-1}=\dfrac{x-\sqrt{x}}{x\sqrt{x}-1}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
b: \(=\dfrac{\sqrt{x}-4+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
c: \(=\dfrac{x\sqrt{x}+1-\left(x-1\right)\left(\sqrt{x}+1\right)}{x-1}\)
\(=\dfrac{x\sqrt{x}+1-x\sqrt{x}-x+\sqrt{x}+1}{x-1}=\dfrac{-x+\sqrt{x}+2}{x-1}\)
\(=\dfrac{-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{x-1}=\dfrac{-\sqrt{x}+2}{\sqrt{x}-1}\)
