Đáp án:
c) x=3
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne - 3\\
\dfrac{{2 - 4x + x\left( {x + 3} \right) - 2\left( {x + 3} \right)}}{{2\left( {x + 3} \right)}} = 0\\
\to 2 - 4x + {x^2} + 3x - 2x - 6 = 0\\
\to {x^2} - 3x - 4 = 0\\
\to \left( {x - 4} \right)\left( {x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 4\\
x = - 1
\end{array} \right.\\
b)DK:x \ge \dfrac{2}{3}\\
\left| {2x - 3} \right| = 3x - 2\\
\to 4{x^2} - 12x + 9 = 9{x^2} - 12x + 4\\
\to 5{x^2} = 5\\
\to \left[ \begin{array}{l}
x = 1\left( {TM} \right)\\
x = - 1\left( l \right)
\end{array} \right.\\
c)DK:x \ge 2\\
\sqrt {2x + 3} + \sqrt {x - 2} = 4\\
\to 2x + 3 + 2\sqrt {\left( {2x + 3} \right)\left( {x - 2} \right)} + x - 2 = 16\\
\to 2\sqrt {\left( {2x + 3} \right)\left( {x - 2} \right)} = - 3x + 15\\
\to 4\left( {2{x^2} - x - 6} \right) = 9{x^2} - 90x + 225\left( {DK:\dfrac{{15}}{3} \ge x} \right)\\
\to {x^2} - 86x + 249 = 0\\
\to \left[ \begin{array}{l}
x = 83\left( l \right)\\
x = 3\left( {TM} \right)
\end{array} \right.
\end{array}\)
