Đáp án:
a) \(\left[ \begin{array}{l}
x = 1\\
x = - 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\sqrt {2{x^2} + 6x + 8} = x + 3\\
\to 2{x^2} + 6x + 8 = {x^2} + 6x + 9\left( {DK:x \ge - 3} \right)\\
\to {x^2} - 1 = 0\\
\to \left( {x - 1} \right)\left( {x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 1\\
x = - 1
\end{array} \right.\left( {TM} \right)\\
b)\sqrt {{x^2} - 5x + 3} = 2x - 1\\
\to {x^2} - 5x + 3 = 4{x^2} - 4x + 1\left( {DK:x \ge \dfrac{1}{2}} \right)\\
\to 3{x^2} + x - 2 = 0\\
\to \left( {3x - 2} \right)\left( {x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{2}{3}\left( {TM} \right)\\
x = - 1\left( l \right)
\end{array} \right.
\end{array}\)
