Đáp án:
B2:
b) \(x = - \dfrac{1}{4}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:DK:a > 0;a \ne 1\\
\left[ {\dfrac{{\sqrt a + 1 + a\sqrt a }}{{a\left( {\sqrt a + 1} \right)}}} \right]:\dfrac{{\sqrt a }}{{\sqrt a \left( {\sqrt a - 1} \right)}}\\
= \dfrac{{a\sqrt a + \sqrt a + 1}}{{a\left( {\sqrt a + 1} \right)}}.\left( {\sqrt a - 1} \right)\\
= \dfrac{{\left( {a\sqrt a + \sqrt a + 1} \right)\left( {\sqrt a - 1} \right)}}{{a\left( {\sqrt a + 1} \right)}}\\
B2:\\
a)\sqrt {{x^3}} = 3\\
\to {x^3} = {3^2}\\
\to {x^3} = 9\\
\to x = \sqrt[3]{9}\\
b)\sqrt {{{\left( {x - 1} \right)}^2}} = 3x + 2\\
\to \left| {x - 1} \right| = 3x + 2\\
\to \left[ \begin{array}{l}
x - 1 = 3x + 2\left( {DK:x \ge 1} \right)\\
x - 1 = - 3x - 2\left( {DK:x < 1} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = - 3\\
4x = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - \dfrac{3}{2}\left( l \right)\\
x = - \dfrac{1}{4}
\end{array} \right.
\end{array}\)
