Đáp án:
a) \(\dfrac{{ - a\sqrt b }}{{\sqrt a + 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:a \ge 0;b \ge 0;a \ne b \ne 1\\
P = \dfrac{{\left( {\sqrt a + 1} \right)\left( {\sqrt {ab} - 1} \right) + \left( {\sqrt {ab} + \sqrt a } \right)\left( {\sqrt {ab} + 1} \right) - ab + 1}}{{ab - 1}}:\dfrac{{\left( {\sqrt a + 1} \right)\left( {\sqrt {ab} - 1} \right) - \left( {\sqrt {ab} + \sqrt a } \right)\left( {\sqrt {ab} + 1} \right) + ab - 1}}{{ab - 1}}\\
= \dfrac{{a\sqrt b - \sqrt a + \sqrt {ab} - 1 + ab + \sqrt {ab} + a\sqrt b + \sqrt a - ab + 1}}{{a\sqrt b - \sqrt a + \sqrt {ab} - 1 - ab - \sqrt {ab} - a\sqrt b - \sqrt a + ab - 1}}\\
= \dfrac{{2a\sqrt b }}{{ - 2\sqrt a - 2}}\\
= \dfrac{{ - a\sqrt b }}{{\sqrt a + 1}}\\
b)Thay:a = 2 - \sqrt 3 \\
\to a = \dfrac{{4 - 2\sqrt 3 }}{2} = \dfrac{{3 - 2\sqrt 3 .1 + 1}}{2} = \dfrac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{2}\\
b = \dfrac{{\sqrt 3 - 1}}{{1 + \sqrt 3 }} = \dfrac{{\left( {\sqrt 3 - 1} \right)\left( {1 - \sqrt 3 } \right)}}{{1 - 3}}\\
= \dfrac{{ - 2 + \sqrt 3 + \sqrt 3 - 1}}{{ - 2}} = \dfrac{{ - 3 + 2\sqrt 3 }}{{ - 2}}\\
= \dfrac{{3 - 2\sqrt 3 }}{2} = \dfrac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{2}\\
\to P = \dfrac{{ - \left( {2 - \sqrt 3 } \right)\sqrt {\dfrac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{2}} }}{{\sqrt {\dfrac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{2}} + 1}}\\
= \dfrac{{ - \left( {2 - \sqrt 3 } \right)\left( {\sqrt 3 - 1} \right)}}{{\sqrt 3 - 1 + 2}}\\
= - 7 + 4\sqrt 3
\end{array}\)
