A=x+2/x
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1 câu trả lời 143
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$\begin{array}{l}
Dkxd:x \ge 0;x \ne 1\\
A = \dfrac{{x + 2}}{{x\sqrt x - 1}} - \dfrac{{\sqrt x + 1}}{{x + \sqrt x + 1}} - \dfrac{1}{{1 - \sqrt x }}\\
= \dfrac{{x + 2}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} - \dfrac{{\sqrt x + 1}}{{x + \sqrt x + 1}} + \dfrac{1}{{\sqrt x - 1}}\\
= \dfrac{{x + 2 - \left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) + x + \sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{x + 2 - \left( {x - 1} \right) + x + \sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{x + 2 - x + 1 + x + \sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{x + \sqrt x + 4}}{{x\sqrt x - 1}}
\end{array}$
Dkxd:x \ge 0;x \ne 1\\
A = \dfrac{{x + 2}}{{x\sqrt x - 1}} - \dfrac{{\sqrt x + 1}}{{x + \sqrt x + 1}} - \dfrac{1}{{1 - \sqrt x }}\\
= \dfrac{{x + 2}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} - \dfrac{{\sqrt x + 1}}{{x + \sqrt x + 1}} + \dfrac{1}{{\sqrt x - 1}}\\
= \dfrac{{x + 2 - \left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) + x + \sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{x + 2 - \left( {x - 1} \right) + x + \sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{x + 2 - x + 1 + x + \sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{x + \sqrt x + 4}}{{x\sqrt x - 1}}
\end{array}$
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