Đáp án:
Giải thích các bước giải:
$\left| {2x + 3} \right| + \left| {2x - 1} \right|{\rm{ }} = \frac{8}{3}{(x + 1)^2} + 2$
+) Nếu $x > \frac{1}{2}$
=> $\begin{array}{l} \left| {2x + 3} \right| = 2x + 3\\ \left| {2x - 1} \right| = 2x - 1 \end{array}$
Khi đó:
$\begin{array}{l} 4x + 2 = \frac{8}{3}{(x + 1)^2} + 2\\ \Leftrightarrow \frac{{8{x^2}}}{3} + \frac{{16x}}{3} + \frac{8}{3} + 2 = 4x + 2\\ \Leftrightarrow \frac{{8{x^2}}}{3} + \frac{{4x}}{3} + \frac{8}{3} = 0(vô\,nghiệm) \end{array}$
+) Nếu $x < \frac{-3}{2}$
=> $\begin{array}{l} \left| {2x + 3} \right| = -2x - 3\\ \left| {2x - 1} \right| = -2x + 1 \end{array}$
Khi đó:
$\begin{array}{l} - 4x - 2 = \frac{8}{3}{(x + 1)^2} + 2\\ \Leftrightarrow \frac{{8{x^2}}}{3} + \frac{{16x}}{3} + \frac{8}{3} + 2 = - 4x - 2\\ \Leftrightarrow \frac{{8{x^2}}}{3} + \frac{{28x}}{3} + \frac{{20}}{3} = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = - \frac{5}{2}\\ x = - 1 \end{array} \right. \end{array}$
(cả 2 trường hợp đều thoả mãn)
+) Nếu $ - \frac{3}{2} \le x \le \frac{1}{2}$
=> $\begin{array}{l} \left| {2x + 3} \right| = 2x + 3\\ \left| {2x - 1} \right| = -2x + 1 \end{array}$
Khi đó:
$\begin{array}{l} - 4 = \frac{8}{3}{(x + 1)^2} + 2\\ \Leftrightarrow \frac{8}{3}{(x + 1)^2} = - 6(vô\,nghiệm) \end{array}$
