Đáp án:
$a = \dfrac{1}{4};b = \dfrac{1}{2}$
Giải thích các bước giải:
ĐKXĐ: $a\ge 0;a\ne 1$
Ta có:
$\begin{array}{l}
\left( {\dfrac{{1 + a\sqrt a }}{{1 + \sqrt a }} - \sqrt a } \right).\dfrac{{a + \sqrt a }}{{1 - a}} = {b^2} - b + \dfrac{1}{2}\\
\Leftrightarrow \left( {\dfrac{{\left( {1 + \sqrt a } \right)\left( {1 - \sqrt a + a} \right)}}{{1 + \sqrt a }} - \sqrt a } \right).\dfrac{{\sqrt a \left( {\sqrt a + 1} \right)}}{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a } \right)}} = {b^2} - b + \dfrac{1}{2}\\
\Leftrightarrow \left( {a - 2\sqrt a + 1} \right).\dfrac{{\sqrt a }}{{\left( {1 - \sqrt a } \right)}} = {b^2} - b + \dfrac{1}{2}\\
\Leftrightarrow {\left( {1 - \sqrt a } \right)^2}.\dfrac{{\sqrt a }}{{\left( {1 - \sqrt a } \right)}} = {b^2} - b + \dfrac{1}{2}\\
\Leftrightarrow \left( {1 - \sqrt a } \right)\sqrt a = {b^2} - b + \dfrac{1}{2}\\
\Leftrightarrow a - \sqrt a + {b^2} - b + \dfrac{1}{2} = 0\\
\Leftrightarrow {\left( {\sqrt a - \dfrac{1}{2}} \right)^2} + {\left( {b - \dfrac{1}{2}} \right)^2} = 0\\
\Leftrightarrow {\left( {\sqrt a - \dfrac{1}{2}} \right)^2} = {\left( {b - \dfrac{1}{2}} \right)^2} = 0\left( {do:{{\left( {\sqrt a - \dfrac{1}{2}} \right)}^2} + {{\left( {b - \dfrac{1}{2}} \right)}^2} \ge 0,\forall a,b} \right)\\
\Leftrightarrow \sqrt a - \dfrac{1}{2} = b - \dfrac{1}{2} = 0\\
\Leftrightarrow a = \dfrac{1}{4};b = \dfrac{1}{2}
\end{array}$
Vậy $a = \dfrac{1}{4};b = \dfrac{1}{2}$
