Bài 7:
a) Ta có: \(G=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right):\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{1-\sqrt{x}}+\dfrac{2}{x-1}\right)\)
\(=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\left(\dfrac{\sqrt{x}-1+\sqrt{x}\left(\sqrt{x}+1\right)+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1+x+\sqrt{x}+2}\)
\(=\dfrac{4\sqrt{x}}{x+2\sqrt{x}+1}\)
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