Đáp án:
a) x=-20
b) \(x = \dfrac{9}{4}\)
c) x=9
d) \(x = - \dfrac{{11}}{4}\)
e) x=2
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:\dfrac{4}{3} \ge x\\
\sqrt {4 - 3x} = 8\\
\to 4 - 3x = 64\\
\to 3x = - 60\\
\to x = - 20\\
b)DK:x \ge 2\\
\sqrt {4\left( {x - 2} \right)} - 12.\dfrac{1}{3}\sqrt {x - 2} = - 1\\
\to 2\sqrt {x - 2} - 4\sqrt {x - 2} = - 1\\
\to - 2\sqrt {x - 2} = - 1\\
\to \sqrt {x - 2} = \dfrac{1}{2}\\
\to x - 2 = \dfrac{1}{4}\\
\to x = \dfrac{9}{4}\\
c)DK:x \ge 0\\
2x - 4\sqrt x + \sqrt x - 2 = 7\\
\to 2x - 3\sqrt x - 9 = 0\\
\to \left( {\sqrt x - 3} \right)\left( {2\sqrt x + 3} \right) = 0\\
\to \sqrt x - 3 = 0\left( {do:2\sqrt x + 3 > 0\forall x \ge 0} \right)\\
\to x = 9\\
d)DK:x \ge - 3\\
\sqrt {25\left( {x + 3} \right)} + 15.\dfrac{1}{5}\sqrt {x + 3} = 2 + 4\sqrt {x + 3} \\
\to 5\sqrt {x + 3} + 3\sqrt {x + 3} - 4\sqrt {x + 3} = 2\\
\to 4\sqrt {x + 3} = 2\\
\to \sqrt {x + 3} = \dfrac{1}{2}\\
\to x + 3 = \dfrac{1}{4}\\
\to x = - \dfrac{{11}}{4}\\
e)\sqrt {{{\left( {x - 1} \right)}^2}} = 2x - 3\\
\to \left| {x - 1} \right| = 2x - 3\\
\to \left[ \begin{array}{l}
x - 1 = 2x - 3\left( {DK:x \ge 1} \right)\\
x - 1 = - 2x + 3\left( {DK:x < 1} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
3x = 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x = \dfrac{4}{3}\left( l \right)
\end{array} \right.
\end{array}\)
