Đáp án:
3) c) \(2\sqrt 3 \)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:x \ge 0;x \ne 1\\
P = \dfrac{{10\sqrt x - \left( {2\sqrt x - 3} \right)\left( {\sqrt x - 1} \right) - \left( {\sqrt x + 1} \right)\left( {\sqrt x + 4} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 4} \right)}}\\
= \dfrac{{10\sqrt x - 2x + 5\sqrt x - 3 - x - 5\sqrt x - 4}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 4} \right)}}\\
= \dfrac{{ - 3x + 10\sqrt x - 7}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 4} \right)}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {7 - 3\sqrt x } \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 4} \right)}}\\
= \dfrac{{7 - 3\sqrt x }}{{\sqrt x + 4}}\\
2)A = \left[ {\dfrac{{3x + 3\sqrt x - \sqrt x + 1 - 3x + 3}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right].\dfrac{{\sqrt x + 1}}{{\sqrt x + 2}}\\
= \dfrac{{2\sqrt x + 4}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{\sqrt x + 1}}{{\sqrt x + 2}}\\
= \dfrac{2}{{\sqrt x - 1}}\\
3)a)A = \dfrac{{4\left( {\sqrt 5 + 1} \right)}}{{5 - 1}} - 2\sqrt 5 - 1\\
= \sqrt 5 + 1 - 2\sqrt 5 - 1 = - \sqrt 5 \\
b)\left( {5.2\sqrt 3 - 4\sqrt 3 + 4\sqrt 3 } \right).\sqrt 3 \\
= 10.3 = 30\\
c)\sqrt 3 - 1 + \sqrt 3 + 1 = 2\sqrt 3
\end{array}\)
