Đáp án:
\(A = x + \sqrt x - 2\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne 1;x \ge 0\\
A = \left( {\dfrac{{\sqrt x + 2}}{{x - 1}} - \dfrac{{\sqrt x + 2}}{{x + 2\sqrt x + 1}}} \right).\dfrac{{{x^2} - 2x + 1}}{2}\\
= \left[ {\dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x + 1} \right) - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right){{\left( {\sqrt x + 1} \right)}^2}}}} \right].\dfrac{{{{\left( {x - 1} \right)}^2}}}{2}\\
= \dfrac{{x + 3\sqrt x + 2 - x - \sqrt x + 2}}{{\left( {\sqrt x - 1} \right){{\left( {\sqrt x + 1} \right)}^2}}}.\dfrac{{{{\left( {x - 1} \right)}^2}}}{2}\\
= \dfrac{{2\sqrt x + 4}}{{\left( {\sqrt x - 1} \right){{\left( {\sqrt x + 1} \right)}^2}}}.\dfrac{{{{\left( {x - 1} \right)}^2}}}{2}\\
= \dfrac{{\sqrt x + 2}}{{\left( {x - 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{{{\left( {x - 1} \right)}^2}}}{1}\\
= \dfrac{{\sqrt x + 2}}{{\sqrt x + 1}}.\left( {x - 1} \right)\\
= \left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)\\
= x + \sqrt x - 2
\end{array}\)
