Đáp án:
\(\dfrac{{{a^2} - \sqrt a }}{{a + \sqrt a + 1}} - \dfrac{{{a^2} + \sqrt a }}{{a - \sqrt a + 1}} + a + 1 = {\left( {\sqrt a - 1} \right)^2}\)
Giải thích các bước giải:
\(\begin{array}{l}
\dfrac{{{a^2} - \sqrt a }}{{a + \sqrt a + 1}} - \dfrac{{{a^2} + \sqrt a }}{{a - \sqrt a + 1}} + a + 1 = {\left( {\sqrt a - 1} \right)^2}\\
VT = \dfrac{{\sqrt a \left( {a\sqrt a - 1} \right)}}{{a + \sqrt a + 1}} - \dfrac{{\sqrt a \left( {a\sqrt a + 1} \right)}}{{a - \sqrt a + 1}} + a + 1\\
= \dfrac{{\sqrt a \left( {\sqrt a - 1} \right)\left( {a + \sqrt a + 1} \right)}}{{a + \sqrt a + 1}} - \dfrac{{\sqrt a \left( {\sqrt a + 1} \right)\left( {a - \sqrt a + 1} \right)}}{{a - \sqrt a + 1}} + a + 1\\
= \sqrt a \left( {\sqrt a - 1} \right) - \sqrt a \left( {\sqrt a + 1} \right) + a + 1\\
= a - \sqrt a - a - \sqrt a + a + 1\\
= a - 2\sqrt a + 1\\
= {\left( {\sqrt a - 1} \right)^2} = VP\\
\to dpcm
\end{array}\)
