Đáp án:
4) -1
1) a) $x = \pm 2\sqrt 2$
c) x = - 2
d) x=6;x=22
Giải thích các bước giải:
\(\begin{array}{l}
4)\, = \sqrt x \sqrt {{{\left( {\sqrt x + 1} \right)}^2}} - \dfrac{{{{\left( {\sqrt x } \right)}^3} - 1}}{{\sqrt x - 1}}\\
= \sqrt x .\left( {\sqrt x + 1} \right) - \dfrac{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}{{\sqrt x - 1}}\\
= x + \sqrt x - x - \sqrt x - 1\\
= - 1\\
1)\,a)\,\sqrt {4 + {x^2}} = 2\sqrt 3 \\
\Leftrightarrow 4 + {x^2} = 12\\
\Leftrightarrow {x^2} = 8\\
\Leftrightarrow x = \pm 2\sqrt 2 \\
c)\, \Leftrightarrow 4x + 1 = - 7\\
\Leftrightarrow 4x = - 8\\
\Leftrightarrow x = - 2\\
d)\,\left( {2x - 3} \right) - 8\sqrt {2x - 3} + 12 = 0\\
\Leftrightarrow \left( {\sqrt {2x - 8} - 2} \right)\left( {\sqrt {2x - 8} - 6} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2x - 8 = 4\\
2x - 8 = 36
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 6\\
x = 22
\end{array} \right.\left( {tm} \right)
\end{array}\)
