Đáp án:
b) \(\left[ \begin{array}{l}
x = - 1\\
x = \dfrac{7}{3}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)2\left| x \right| = 2 + \left| {x + 1} \right|\\
\to 4{x^2} = 4 + 4\left( {x + 1} \right) + {x^2} + 2x + 1\\
\to 3{x^2} - 6x - 9 = 0\\
\to {x^2} - 2x - 9 = 0\\
\to \left[ \begin{array}{l}
x = - 3\\
x = - 1
\end{array} \right.\\
b)\left| {2x + 1} \right| + \left| {x - 3} \right| = 5\\
\to 4{x^2} + 4x + 1 + 2\left( {2x + 1} \right)\left( {x - 3} \right) + {x^2} - 6x + 9 = 25\\
\to 5{x^2} - 2x - 15 + 2\left( {2{x^2} - 5x - 3} \right) = 0\\
\to 9{x^2} - 12x - 21 = 0\\
\to 3{x^2} - 4x - 7 = 0\\
\to \left[ \begin{array}{l}
x = - 1\\
x = \dfrac{7}{3}
\end{array} \right.
\end{array}\)
