Đáp án:
\(D = {\left( {\sqrt a + \sqrt b } \right)^2}\)
Giải thích các bước giải:
\(\begin{array}{l}
A = \left( {\dfrac{{1 - a\sqrt a }}{{1 - \sqrt a }} + \sqrt a } \right){\left( {\dfrac{{1 - \sqrt a }}{{1 - a}}} \right)^2}\\
= \left[ {\dfrac{{\left( {1 - \sqrt a } \right)\left( {a + \sqrt a + 1} \right)}}{{1 - \sqrt a }} + \sqrt a } \right].{\left[ {\dfrac{{1 - \sqrt a }}{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a } \right)}}} \right]^2}\\
= \left( {a + \sqrt a + 1 + \sqrt a } \right).\dfrac{1}{{1 + \sqrt a }}\\
= {\left( {\sqrt a + 1} \right)^2}.\dfrac{1}{{1 + \sqrt a }}\\
= \sqrt a + 1\\
B = \left( {\dfrac{{1 - a\sqrt a }}{{1 - \sqrt a }} + \sqrt a } \right)\left( {\dfrac{{1 + a\sqrt a }}{{1 + \sqrt a }} - \sqrt a } \right)\\
= \left[ {\dfrac{{\left( {1 - \sqrt a } \right)\left( {a + \sqrt a + 1} \right)}}{{1 - \sqrt a }} + \sqrt a } \right].\left[ {\dfrac{{\left( {1 + \sqrt a } \right)\left( {a - \sqrt a + 1} \right)}}{{1 + \sqrt a }} - \sqrt a } \right]\\
= \left( {a + \sqrt a + 1 + \sqrt a } \right)\left( {a - \sqrt a + 1 - \sqrt a } \right)\\
= {\left( {\sqrt a + 1} \right)^2}{\left( {\sqrt a - 1} \right)^2}\\
= {\left[ {\left( {\sqrt a + 1} \right)\left( {\sqrt a - 1} \right)} \right]^2}\\
= {\left( {a - 1} \right)^2}\\
C = \dfrac{{a\sqrt b - b\sqrt a }}{{\sqrt {ab} }}:\left( {\sqrt a - \sqrt b } \right)\\
= \dfrac{{\sqrt {ab} \left( {\sqrt a - \sqrt b } \right)}}{{\sqrt {ab} }}.\dfrac{1}{{\sqrt a - \sqrt b }}\\
= 1\\
D = \dfrac{{a\sqrt a - b\sqrt b }}{{\sqrt a - \sqrt b }} + \sqrt {ab} \\
= \dfrac{{\left( {\sqrt a - \sqrt b } \right)\left( {a + \sqrt {ab} + b} \right)}}{{\sqrt a - \sqrt b }} + \sqrt {ab} \\
= a + \sqrt {ab} + b + \sqrt {ab} \\
= {\left( {\sqrt a + \sqrt b } \right)^2}
\end{array}\)
