Đáp án:
\(\left[ \begin{array}{l}
x = \dfrac{{ - 3 + \sqrt 6 }}{3}\\
x = - 1,579665221
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\left| {2x + 3} \right| + \left| {2x - 1} \right| = \dfrac{8}{{3{{\left( {x + 1} \right)}^2} + 2}}\\
\to \left| {2x + 3} \right| + \left| {2x - 1} \right| = \dfrac{8}{{3{x^2} + 6x + 3 + 2}}\\
\to \left| {2x + 3} \right| + \left| {2x - 1} \right| = \dfrac{8}{{3{x^2} + 6x + 5}}\\
TH1:x \ge \dfrac{1}{2}\\
Pt \to 2x + 3 + 2x - 1 = \dfrac{8}{{3{x^2} + 6x + 5}}\\
\to 4x + 2 = \dfrac{8}{{3{x^2} + 6x + 5}}\\
\to 2x + 1 = \dfrac{4}{{3{x^2} + 6x + 5}}\\
\to 6{x^3} + 12{x^2} + 10x + 3{x^2} + 6x + 5 = 4\\
\to 6{x^3} + 15{x^2} + 16x + 1 = 0\\
\to x = - 0,06654041966\left( l \right)\\
TH2:\dfrac{1}{2} > x \ge - \dfrac{3}{2}\\
Pt \to 2x + 3 - 2x + 1 = \dfrac{8}{{3{x^2} + 6x + 5}}\\
\to 4 = \dfrac{8}{{3{x^2} + 6x + 5}}\\
\to 1 = \dfrac{4}{{3{x^2} + 6x + 5}}\\
\to 3{x^2} + 6x + 5 = 4\\
\to 3{x^2} + 6x + 1 = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{{ - 3 + \sqrt 6 }}{3}\left( {TM} \right)\\
x = \dfrac{{ - 3 - \sqrt 6 }}{3}\left( l \right)
\end{array} \right.\\
TH3: - \dfrac{3}{2} > x\\
Pt \to - 2x - 3 - 2x + 1 = \dfrac{8}{{3{x^2} + 6x + 5}}\\
\to - 2 - 4x = \dfrac{8}{{3{x^2} + 6x + 5}}\\
\to - 1 - 2x = \dfrac{4}{{3{x^2} + 6x + 5}}\\
\to 2x + 1 = \dfrac{{ - 4}}{{3{x^2} + 6x + 5}}\\
\to 6{x^3} + 12{x^2} + 10x + 3{x^2} + 6x + 5 = - 4\\
\to 6{x^3} + 16{x^2} + 16x + 9 = 0\\
\to x = - 1,579665221\left( {TM} \right)
\end{array}\)
\(KL:\left[ \begin{array}{l}
x = \dfrac{{ - 3 + \sqrt 6 }}{3}\\
x = - 1,579665221
\end{array} \right.\)
