Đáp án:
\(\begin{array}{l}
a)A = \dfrac{1}{{x - 1}}\\
b)5\sqrt 3 \\
c)7\\
d)C = \dfrac{{2a}}{{a - b}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \left[ {\dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right) + \left( {\sqrt x + 2} \right)\left( {1 - \sqrt x } \right)}}{{{{\left( {\sqrt x + 1} \right)}^2}\left( {1 - \sqrt x } \right)}}} \right].\dfrac{{\sqrt x + 1}}{{2\sqrt x }}\\
= \dfrac{{x - \sqrt x - 2 - x - \sqrt x + 2}}{{{{\left( {\sqrt x + 1} \right)}^2}\left( {1 - \sqrt x } \right)}}.\dfrac{{\sqrt x + 1}}{{2\sqrt x }}\\
= \dfrac{{ - 2\sqrt x }}{{{{\left( {\sqrt x + 1} \right)}^2}\left( {1 - \sqrt x } \right)}}.\dfrac{{\sqrt x + 1}}{{2\sqrt x }}\\
= - \dfrac{1}{{1 - x}} = \dfrac{1}{{x - 1}}\\
b)A = 6.2\sqrt 3 - 5.3\sqrt 3 + 2.4\sqrt 3 \\
= \left( {12 - 15 + 8} \right)\sqrt 3 = 5\sqrt 3 \\
c)B = \sqrt {4 - 2.2.\sqrt 3 + 3} + \sqrt {25 + 2.5.\sqrt 3 + 3} \\
= \sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} + \sqrt {{{\left( {5 + \sqrt 3 } \right)}^2}} \\
= 2 - \sqrt 3 + 5 + \sqrt 3 = 7\\
d)DK:a \ge 0;b \ge 0;a \ne b\\
C = \dfrac{{\sqrt a \left( {\sqrt a + \sqrt b } \right) + \sqrt a \left( {\sqrt a - \sqrt b } \right)}}{{\left( {\sqrt a + \sqrt b } \right)\left( {\sqrt a - \sqrt b } \right)}}\\
= \dfrac{{a + \sqrt {ab} + a - \sqrt {ab} }}{{a - b}} = \dfrac{{2a}}{{a - b}}
\end{array}\)
