thực hiện phép tính (1-a ²) :[ ($\frac{1-a√a}{1- √a}$ + √a) . ($\frac{1+ a√a}{1+√a}$ - √a)]+1
1 câu trả lời
Đáp án:
$\begin{array}{l}
\left( {1 - {a^2}} \right):\left[ {\left( {\dfrac{{1 - a\sqrt a }}{{1 - \sqrt a }} + \sqrt a } \right).\left( {\dfrac{{1 + a\sqrt a }}{{1 + \sqrt a }} - \sqrt a } \right)} \right] + 1\\
= \left( {1 + a} \right)\left( {1 - a} \right)\\
:\left[ {\left( {\dfrac{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a + a} \right)}}{{1 - \sqrt a }} + \sqrt a } \right).\left( {\dfrac{{\left( {1 + \sqrt a } \right)\left( {1 - \sqrt a + a} \right)}}{{1 + \sqrt a }} - \sqrt a } \right)} \right]\\
+ 1\\
= \left( {1 + a} \right)\left( {1 - a} \right)\\
:\left[ {\left( {1 + \sqrt a + a + \sqrt a } \right)\left( {1 - \sqrt a + a - \sqrt a } \right)} \right] + 1\\
= \left( {1 + a} \right)\left( {1 - a} \right):\left[ {{{\left( {1 + \sqrt a } \right)}^2}.{{\left( {1 - \sqrt a } \right)}^2}} \right] + 1\\
= \left( {1 + a} \right)\left( {1 - a} \right):{\left( {1 - a} \right)^2} + 1\\
= \dfrac{{1 + a}}{{1 - a}} + 1\\
= \dfrac{{1 + a + 1 - a}}{{1 - a}}\\
= \dfrac{2}{{1 - a}}
\end{array}$