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\(A=\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x}-1}\right)\)\(:\left(\frac{\sqrt{x}+2}{\sqrt{x}-1}-\frac{\sqrt{x}+1}{\sqrt{x}-2}\right)\)
\(=\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\)\(:\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{\left(\sqrt{x}-1-\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\left(\sqrt{x}-4-\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{-3}\)\(=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
\(b,A=0\Leftrightarrow\frac{\sqrt{x}-2}{3\sqrt{x}}=0\Leftrightarrow\sqrt{x}-2=0\)
Mà \(\sqrt{x}+2\ne0\)\(\Rightarrow\)không có giá trị nào của x thỏa mãn \(A=0\)
a) Ta có: \(A=\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right)\cdot\left(\frac{1-\sqrt{x}}{1-x}\right)^2\)
\(=\left(\frac{1-x\sqrt{x}+\sqrt{x}\left(1-\sqrt{x}\right)}{1-\sqrt{x}}\right)\cdot\left(\frac{1}{1+\sqrt{x}}\right)^2\)
\(=\frac{1-x\sqrt{x}+\sqrt{x}-x}{1-\sqrt{x}}\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{-\left(x-1\right)\left(-1-\sqrt{x}\right)}{1-\sqrt{x}}\cdot\frac{1}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{\left(1+\sqrt{x}\right)\cdot\left(-1-\sqrt{x}\right)}{\left(1+\sqrt{x}\right)^2}\)
\(=\frac{-1\cdot\left(1+\sqrt{x}\right)^2}{\left(1+\sqrt{x}\right)^2}=-1\)
a)
\(A=\left(\frac{x-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\frac{x-1}{x-2\sqrt{x}}\\ =\left(\frac{x-1-4\sqrt{x}+\sqrt{x}+1}{x-1}\right)\cdot\frac{x-1}{x-2\sqrt{x}}\\ =\frac{x-3\sqrt{x}}{x-1}\cdot\frac{x-1}{x-2\sqrt{x}}\\ =\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\frac{\sqrt{x}-3}{\sqrt{x}-2}\)
b)
\(A=\frac{\sqrt{x}-3}{\sqrt{x}-2}=\frac{1}{2}\\ \Leftrightarrow2\left(\sqrt{x}-3\right)=\sqrt{x}-2\\ \Leftrightarrow2\sqrt{x}-6=\sqrt{x}-2\\ \Leftrightarrow\sqrt{x}=4\\ \Leftrightarrow x=16\left(tm\right)\)
Điều kiện: \(\hept{\begin{cases}x>0;x\ne1;x\ne4\\\sqrt{x}-1>0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne4\\x>1\end{cases}}}\)
Để A dương <=>\(2-\sqrt{x}>0\Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)
Đối chiếu điều kiện ta có: 1<x<4