Đáp án:
a) x=5
b) x=64
c) x=3
d) x=3
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge - \dfrac{1}{3}\\
\sqrt {3x + 1} = 4\\
\to 3x + 1 = 16\\
\to 3x = 15\\
\to x = 5\\
c)\sqrt {4{x^2} + 4x + 1} = 3x - 2\\
\to 4{x^2} + 4x + 1 = 9{x^2} - 12x + 4\left( {DK:x \ge \dfrac{2}{3}} \right)\\
\to 5{x^2} - 16x + 3 = 0\\
\to \left( {x - 3} \right)\left( {5x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 3\\
x = \dfrac{1}{5}\left( l \right)
\end{array} \right.\\
b)DK:x \ge 0\\
2.3\sqrt x - 2.5\sqrt x + 3.2\sqrt x = 16\\
\to 2\sqrt x = 16\\
\to \sqrt x = 8\\
\to x = 64\\
d)DK:x \ge 2\\
\sqrt {x + 1} = 1 + \sqrt {x - 2} \\
\to x + 1 = 1 + 2\sqrt {x - 2} + x - 2\\
\to 2\sqrt {x - 2} = 2\\
\to \sqrt {x - 2} = 1\\
\to x - 2 = 1\\
\to x = 3
\end{array}\)
