Giải thích các bước giải:
\(\begin{array}{l}
2)a) \Leftrightarrow 3x + \sqrt 2 = 2x + 2\sqrt 2 \\
\Leftrightarrow 3x - 2x = 2\sqrt 2 - \sqrt 2 \\
\Leftrightarrow x = \sqrt 2 \\
b)DK:x \ge 2\\
3\sqrt {x - 2} = \sqrt {{x^2} - 4} \\
\Leftrightarrow 9\left( {x - 2} \right) = {x^2} - 4\\
\Leftrightarrow \left( {x - 2} \right)\left( {x + 2 - 9} \right) = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {x - 7} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = 7
\end{array} \right.\\
3a)\sqrt[3]{{ - 8}} - \sqrt {100} = - 2 - 10 = - 12\\
b) = \left| {\sqrt 2 - 3} \right| + \sqrt 2 - 5\\
= 3 - \sqrt 2 + \sqrt 2 - 5 = - 2\\
4)a)DK:a \ge 0;a \ne 1\\
b)P = \dfrac{{ - 2{a^2} - 4}}{{\left( {a - 1} \right)\left( {{a^2} + a + 1} \right)}} - \dfrac{{\sqrt a - 1}}{{a - 1}} + \dfrac{{\sqrt a + 1}}{{a - 1}}\\
= \dfrac{{ - 2{a^2} - 4}}{{\left( {a - 1} \right)\left( {{a^2} + a + 1} \right)}} + \dfrac{2}{{a - 1}}\\
= \dfrac{{ - 2{a^2} - 4 + 2\left( {{a^2} + a + 1} \right)}}{{\left( {a - 1} \right)\left( {{a^2} + a + 1} \right)}}\\
= \dfrac{{2a - 2}}{{\left( {a - 1} \right)\left( {{a^2} + a + 1} \right)}}\\
= \dfrac{{2\left( {a - 1} \right)}}{{\left( {a - 1} \right)\left( {{a^2} + a + 1} \right)}}\\
= \dfrac{2}{{\left( {a - 1} \right)\left( {{a^2} + a + 1} \right)}}
\end{array}\)
