a) Ta có: \(X=\dfrac{3}{\sqrt{x-3}-\sqrt{x}}+\dfrac{3}{\sqrt{x-3}+\sqrt{x}}+\dfrac{x\sqrt{x}+x}{\sqrt{x}+1}\)
\(=\dfrac{3\sqrt{x-3}+3\sqrt{x}+3\sqrt{x-3}-3\sqrt{x}}{x-3-x}+\dfrac{x\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\dfrac{6\sqrt{x-3}}{-3}+x\)
\(=x-2\sqrt{x-3}\)
b) Để X>2 thì x-2>0
\(\Leftrightarrow x-2\sqrt{x-3}-2>0\)
\(\Leftrightarrow x-3-2\cdot\sqrt{x-3}\cdot1+1>0\)
\(\Leftrightarrow\left(\sqrt{x-3}-1\right)^2>0\)
\(\Leftrightarrow\sqrt{x-3}>1\)
\(\Leftrightarrow x-3>1\)
hay x>4
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