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a) Ta có: \(A=\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right)\cdot\left(\frac{1-\sqrt{x}}{1-x}\right)^2\)
\(=\left(\frac{1-x\sqrt{x}+\sqrt{x}\left(1-\sqrt{x}\right)}{1-\sqrt{x}}\right)\cdot\left(\frac{1}{1+\sqrt{x}}\right)^2\)
\(=\frac{1-x\sqrt{x}+\sqrt{x}-x}{1-\sqrt{x}}\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{-\left(x-1\right)\left(-1-\sqrt{x}\right)}{1-\sqrt{x}}\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{\left(1+\sqrt{x}\right)\cdot\left(-1-\sqrt{x}\right)}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{-1\cdot\left(1+\sqrt{x}\right)^2}{\left(1+\sqrt{x}\right)^2}=-1\)
B=\(\frac{x\sqrt{x}-1}{x-1}\)(x>0,x≠1)
=\(\frac{\sqrt{x^3}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\left(\frac{x+2-\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right):\left(\frac{\left(\sqrt{x}-4\right)\left(x+1\right)-\sqrt{x}\left(1-x\right)}{1-x^2}\right)\)
\(=\left(\frac{x+2-x-\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{x\sqrt{x}+\sqrt{x}-4x-4-\sqrt{x}+x\sqrt{x}}{1-x^2}\right)\)
\(=\frac{2-\sqrt{x}}{\sqrt{x}+1}:\frac{2x\sqrt{x}-3x-4}{\left(1-x\right)\left(1+x\right)}\)
\(=\frac{2-\sqrt{x}}{\sqrt{x}+1}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(1+x\right)}{2x\sqrt{x}-3x-4}\)
\(=\frac{\left(2-\sqrt{x}\right)\left(\sqrt{x}+x\sqrt{x}-1+x\right)}{2x\sqrt{x}-3x-4}\)
\(=\frac{2\sqrt{x}+2x\sqrt{x}-2+2x-x-x^2+\sqrt{x}-x\sqrt{x}}{2x\sqrt{x}-3x-4}\)
tới đêy tự xử đi
a) đk : \(x\ge0\) ; \(x\ne1\)
A=\(\left(\frac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}+1\right)}-\frac{x+1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
\(=\left(\frac{-\left(\sqrt{x}-1\right)^2}{\left(x+1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\) \(=\frac{1-\sqrt{x}}{x+1}\)
b) đk : \(x\ne0;x\ne1\)
B=\(\left(\frac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2}{x-1}\right):\left(\frac{1-x}{2\sqrt{x}}\right)^2\) \(=\left(\frac{-2\sqrt{x}}{x-1}\right):\left(\frac{1-x}{2\sqrt{x}}\right)^2\) \(=\frac{-4x}{\left(x-1\right)^3}\)
\(A=\left(\sqrt{x}-\frac{x+2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-4}{1-x}\right)\) \(ĐKXĐ:x\ge0;x\ne1;x\ne4\)
\(A=\left[\frac{\sqrt{x}\left(\sqrt{x}+1\right)-x-2}{\sqrt{x}+1}\right]:\left[\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}-4}{x-1}\right]\)
\(A=\frac{x+\sqrt{x}-x-2}{\sqrt{x}+1}:\left[\frac{x-\sqrt{x}+\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\)
\(A=\frac{\sqrt{x}-2}{\sqrt{x}+1}:\frac{x-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{\sqrt{x}-2}{\sqrt{x}+1}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(A=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
vậy \(A=\frac{\sqrt{x}-1}{\sqrt{x}+2}\)
b)theo bài ra: \(A=\frac{1}{\sqrt{x}}\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{1}{\sqrt{x}}\)
\(\Leftrightarrow\left(\sqrt{x}-1\right).\sqrt{x}=\sqrt{x}+2\)
\(\Leftrightarrow x-\sqrt{x}-\sqrt{x}-2=0\)
\(\Leftrightarrow x-2\sqrt{x}-2=0\)
\(\Leftrightarrow x-2\sqrt{x}+1-3=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2-\left(\sqrt{3}\right)^2=0\)
\(\Leftrightarrow\left(\sqrt{x}-1-\sqrt{3}\right)\left(\sqrt{x}-1+\sqrt{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-1-\sqrt{3}=0\\\sqrt{x}-1+\sqrt{3}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=\sqrt{3}+1\\\sqrt{x}=1-\sqrt{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\left(\sqrt{3}+1\right)^2\\x=\left(1-\sqrt{3}\right)^2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3+2\sqrt{3}+1\\x=3-2\sqrt{3}+1\end{cases}}\)
vậy......
đề bài khỏi chép
\(A=\left[\frac{x-\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\frac{x}{\sqrt{x}\left(\sqrt{x}-2\right)}\right]:\frac{\sqrt{x}-1}{\sqrt{x}-2}\)
\(A=\left[\frac{x-\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\right]\) \(:\frac{\sqrt{x}-1}{\sqrt{x}-2}\)
\(A=\frac{x-\sqrt{x}+2-x-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(A=\frac{-2\sqrt{x}+2}{\sqrt{x}+1}.\frac{1}{\sqrt{x}-1}\)
\(A=\frac{-2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\frac{-2}{\sqrt{x}+1}\)
vậy....