a) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)
b) Ta có: \(P=\left(\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right):\frac{2x}{x-4}\)
\(=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\cdot\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2x}\)
\(=\frac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2x}\)
\(=\frac{1}{\sqrt{x}}\)
c) Để P<1 thì P-1<0
\(\Leftrightarrow\frac{1}{\sqrt{x}}-1< 0\)
\(\Leftrightarrow\frac{1-\sqrt{x}}{\sqrt{x}}< 0\)
mà \(\sqrt{x}>0\forall x\) thỏa mãn ĐKXĐ
nên \(1-\sqrt{x}< 0\)
\(\Leftrightarrow\sqrt{x}>1\)
hay x>1
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x\ne4\\x>1\end{matrix}\right.\)
Vậy: Để P<1 thì \(\left\{{}\begin{matrix}x\ne4\\x>1\end{matrix}\right.\)
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