## ( √x+2/x-2 √x+1- √x-2/x-1) × √x+1/ √x với x>0

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( √x+2/x-2 √x+1- √x-2/x-1) × √x+1/ √x với x>0

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1 year 2020-11-27T06:33:51+00:00 1 Answers 40 views 0

$\begin{array}{l} x > 0\\ A = \left( {\dfrac{{\sqrt x + 2}}{{x – 2\sqrt x + 1}} – \dfrac{{\sqrt x – 2}}{{x – 1}}} \right).\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\ = \left[ {\dfrac{{\sqrt x + 2}}{{{{\left( {\sqrt x – 1} \right)}^2}}} – \dfrac{{\sqrt x – 2}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}} \right].\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\ = \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x + 1} \right) – \left( {\sqrt x – 2} \right)\left( {\sqrt x – 1} \right)}}{{{{\left( {\sqrt x – 1} \right)}^2}.\left( {\sqrt x + 1} \right)}}.\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\ = \dfrac{{x + 3\sqrt x + 2 – \left( {\sqrt x – 3\sqrt x + 2} \right)}}{{{{\left( {\sqrt x – 1} \right)}^2}}}.\dfrac{1}{{\sqrt x }}\\ = \dfrac{{6\sqrt x }}{{{{\left( {\sqrt x – 1} \right)}^2}.\sqrt x }}\\ = \dfrac{6}{{{{\left( {\sqrt x – 1} \right)}^2}}} \end{array}$