Đáp án:
\(\dfrac{{ - 4x + 9\sqrt x - 2}}{{x + \sqrt x - 2}}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0;x \ne 4\\
A = \dfrac{{\left( {x + 2} \right)\left( {\sqrt x - 2} \right) - 2\sqrt x \left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) - \left( {1 - \sqrt x } \right)\left( {x - \sqrt x - 2} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x\sqrt x - 2x + 2\sqrt x - 4 - 2\sqrt x \left( {x - 4} \right) - x + \sqrt x + 2 + x\sqrt x - x - 2\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{2x\sqrt x - 4x + \sqrt x - 2 - 2x\sqrt x + 8\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{ - 4x + 9\sqrt x - 2}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\left( {2 - \sqrt x } \right)\left( {4\sqrt x - 1} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{1 - 4\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}\\
B = \dfrac{{\sqrt x - 2 - \sqrt x + 3}}{{\sqrt x - 2}}\\
= \dfrac{1}{{\sqrt x - 2}}\\
C = A:B\\
= \dfrac{{1 - 4\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}:\dfrac{1}{{\sqrt x - 2}}\\
= \dfrac{{\left( {1 - 4\sqrt x } \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{ - 4x + 9\sqrt x - 2}}{{x + \sqrt x - 2}}
\end{array}\)
