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![](https://rs.olm.vn/images/avt/0.png?1311)
ĐK \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
Ta có \(A=\left(\frac{1}{\sqrt{x}-1}+\frac{x-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}+2}-\frac{x-\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right)\)
\(=\frac{\sqrt{x}+2+x-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}:\frac{x-1-x+\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+3}=\frac{x+3}{\sqrt{x}+3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A=\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A=\left(\dfrac{\sqrt{3}\left(x-\sqrt{3}\right)+3}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\right)\cdot\dfrac{x^2+3+x\sqrt{3}}{x\sqrt{3}}\)
\(=\dfrac{x\sqrt{3}}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\cdot\dfrac{x^2+x\sqrt{3}+3}{x\sqrt{3}}\)
\(=\dfrac{1}{x-\sqrt{3}}\)
b: \(B=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\)
\(=x-\sqrt{x}-x-\sqrt{x}+x+1\)
\(=x-2\sqrt{x}+1\)
c: \(C=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{x\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2-\left(x-\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(P=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b:Sửa đề: 2A
2A=2căn x+5
=>(2căn x+2)/căn x=2căn x+5
=>2x+5căn x-2căn x-2=0
=>2x+3căn x-2=0
=>(căn x+2)(2căn x-1)=0
=>x=1/4
\(A=1-\left(\dfrac{2}{1+2\sqrt{x}}-\dfrac{5\sqrt{x}}{4x-1}-\dfrac{1}{1-2\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{4x+4\sqrt{x}+1}\)
\(=1-\dfrac{2\left(4x-1\right)-\left(1-2\sqrt{x}\right)-5\sqrt{x}\cdot\left(1+2\sqrt{x}\right)\cdot\left(1-2\sqrt{x}\right)-\left(1-2\sqrt{x}\right)\cdot\left(4x-1\right)}{\left(1+2\sqrt{x}\right)\cdot\left(4x-1\right)\cdot\left(1-2\sqrt{x}\right)}\cdot\dfrac{4x+4\sqrt{x}+1}{\sqrt{x}-1}\)
\(=1-\dfrac{4x-4x\sqrt{x}-1+\sqrt{x}}{\left(1+2\sqrt{x}\right)\cdot\left(4x-1\right)\cdot\left(1-2\sqrt{x}\right)}\cdot\dfrac{4x+4\sqrt{x}+1}{\sqrt{x}-1}\)
\(=1-\dfrac{4x\cdot\left(1-\sqrt{x}\right)-\left(1-\sqrt{x}\right)}{\left(1+2\sqrt{x}\right)\cdot\left(4x-1\right)\cdot\left(1-2\sqrt{x}\right)}\cdot\dfrac{4x+4\sqrt{x}+1}{\sqrt{x}-1}\)
\(=1-\dfrac{\left(4x-1\right)\cdot\left(1-\sqrt{x}\right)}{\left(1+2\sqrt{x}\right)\cdot\left(4x-1\right)\cdot\left(1-2\sqrt{x}\right)}\cdot\dfrac{4x+4\sqrt{x}+1}{\sqrt{x}-1}\)
\(=1-\dfrac{1-\sqrt{x}}{\left(1+2\sqrt{x}\right)\cdot\left(1-2\sqrt{x}\right)}\cdot\dfrac{4x+4\sqrt{x}+1}{\sqrt{x}-1}\)
\(=1-\dfrac{-\left(\sqrt{x}-1\right)}{\left(1+2\sqrt{x}\right)\cdot\left(1-2\sqrt{x}\right)}\cdot\dfrac{4x+4\sqrt{x}+1}{\sqrt{x}-1}\)
\(=1-\dfrac{-1}{\left(1-2\sqrt{x}\right)\cdot\left(1-2\sqrt{x}\right)}\cdot\left(4x+4\sqrt{x}+1\right)\)
\(=1+\dfrac{1}{1-4x}\cdot\left(4x+4\sqrt{x}+1\right)\)
\(=1+\dfrac{4x+4\sqrt{x}+1}{1-4x}\)
\(=\dfrac{1-4x+4x+4\sqrt{x}+1}{1-4x}\)
\(=\dfrac{2+4\sqrt{x}}{1-4x}\)
kết quả chưa tối giản thế này mới đúng
\(\dfrac{2}{1-2\sqrt{x}}\)