Đáp án:
\(\begin{array}{l}
20)A = \dfrac{2}{{\sqrt x - 1}}\\
21)A = 6\\
22)P = \dfrac{{4\sqrt x }}{{3 + \sqrt x }}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
20)A = \dfrac{{3\sqrt x \left( {\sqrt x + 1} \right) - \sqrt x + 1 - 3\left( {x - 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{{\sqrt x + 1}}{{\sqrt x + 2}}\\
= \dfrac{{3x + 3\sqrt x - \sqrt x + 1 - 3x + 3}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{{\sqrt x + 1}}{{\sqrt x + 2}}\\
= \dfrac{{2\sqrt x + 4}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{{\sqrt x + 1}}{{\sqrt x + 2}}\\
= \dfrac{2}{{\sqrt x - 1}}\\
21)A = \left[ {\dfrac{{\sqrt a \left( {\sqrt a + 1} \right)}}{{\sqrt a + 1}} + \dfrac{{{{\left( {\sqrt a - 2} \right)}^2}}}{{\sqrt a - 2}}} \right].\dfrac{3}{{\sqrt a - 1}}\\
= \left( {\sqrt a + \sqrt a - 2} \right).\dfrac{3}{{\sqrt a - 1}}\\
= \left( {2\sqrt a - 2} \right).\dfrac{3}{{\sqrt a - 1}}\\
= 6\\
22)P = \dfrac{{4x + {{\left( {2 + \sqrt x } \right)}^2} - {{\left( {2 - \sqrt x } \right)}^2}}}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}.\dfrac{{2 - \sqrt x }}{{3 + \sqrt x }}\\
= \dfrac{{4x + 4 + 4\sqrt x + x - 4 + 4\sqrt x - x}}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}.\dfrac{{2 - \sqrt x }}{{3 + \sqrt x }}\\
= \dfrac{{4x + 8\sqrt x }}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}.\dfrac{{2 - \sqrt x }}{{3 + \sqrt x }}\\
= \dfrac{{4\sqrt x }}{{3 + \sqrt x }}
\end{array}\)
