Toán Lớp 9: thực hiện phép tính
(1-a ²) :[ ($\frac{1-a√a}{1- √a}$ + √a) . ($\frac{1+ a√a}{1+√a}$ – √a)]+1
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Toán Lớp 9: thực hiện phép tính
(1-a ²) :[ ($\frac{1-a√a}{1- √a}$ + √a) . ($\frac{1+ a√a}{1+√a}$ – √a)]+1
Comments ( 1 )
\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}$