a) Ta có: \(A=\dfrac{1}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{4-x}\)
\(=\dfrac{2-\sqrt{x}+2+\sqrt{x}-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(=\dfrac{4-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(=\dfrac{2}{\sqrt{x}+2}\)
b) Để \(A=\dfrac{1}{4}\) thì \(\dfrac{2}{\sqrt{x}+2}=\dfrac{1}{4}\)
\(\Leftrightarrow\sqrt{x}+2=8\)
\(\Leftrightarrow\sqrt{x}=6\)
hay x=36(thỏa ĐK)
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