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a. ĐKXĐ: \(x\ne0;1\)
Ta có: \(A=\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{x\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{x\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\left(x\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+\left(x+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x^2+2x\sqrt{x}-2\sqrt{x}-1}{\sqrt{x}\left(x-1\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)^2\left(x-1\right)}{\sqrt{x}\left(x-1\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
b. \(A=4\Leftrightarrow4=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)^2=4\sqrt{x}\)
\(\Leftrightarrow x+1+2\sqrt{x}-4\sqrt{x}=0\)
\(\Leftrightarrow x+1-2\sqrt{x}=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2=0\)
\(\Leftrightarrow\sqrt{x}=1\)
\(\Leftrightarrow x=1\)
Vậy....
\(A=\left(\dfrac{1}{\sqrt{x-1}}+\dfrac{1}{\sqrt{x-1}}\right)^2\cdot\dfrac{x^2-1}{2}-\sqrt{x^2-1}\) (ĐK: \(x>1\))
\(A=\left(\dfrac{2}{\sqrt{x-1}}\right)^2\cdot\dfrac{x^2-1}{2}-\sqrt{x^2-1}\)
\(A=\dfrac{4}{x-1}\cdot\dfrac{\left(x+1\right)\left(x-1\right)}{2}-\sqrt{x^2-1}\)
\(A=2\left(x+1\right)-\sqrt{\left(x+1\right)\left(x-1\right)}\)
\(A=\sqrt{x+1}\left(2\sqrt{x+1}-\sqrt{x-1}\right)\)
\(A=\left(\dfrac{1}{\sqrt{x-1}}+\dfrac{1}{\sqrt{x+1}}\right)^2\cdot\dfrac{x^2-1}{2}-\sqrt{x^2-1}\\ \Rightarrow A=\left(\dfrac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x^2-1}}\right)^2\cdot\dfrac{x^2-1}{2}-\sqrt{x^2-1}\\ \Rightarrow A=\dfrac{\left(\sqrt{x+1}+\sqrt{x-1}\right)^2}{2}-\sqrt{x^2-1}\\ \Rightarrow A=\dfrac{2x+2\sqrt{x^2-1}-2\sqrt{x^2-1}}{2}\\ \Rightarrow A=x\)