Đáp án:
b) \(x = - \dfrac{1}{4}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{{3x + 2}}{2} - \dfrac{{3x + 1}}{6} \ge 2x + \dfrac{5}{3}\\
\to \dfrac{{9x + 6 - 3x - 1 - 6.2x - 10}}{6} \ge 0\\
\to 9x + 6 - 3x - 1 - 12x - 10 \ge 0\\
\to - 6x - 5 \ge 0\\
\to - 6x \ge 5\\
\to x \le - \dfrac{5}{6}\\
c)3x + 1 + \left| {x - 4} \right| = 4\\
\to \left| {x - 4} \right| = - 3x + 3\\
\to \left[ \begin{array}{l}
x - 4 = - 3x + 3\\
x - 4 = 3x - 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
4x = 7\\
2x = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{7}{4}\\
x = - \dfrac{1}{2}
\end{array} \right.\\
b)DK:x \ne 1\\
\dfrac{{{x^2} + x + 1 - 3{x^2} - 2x\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = 0\\
\to - 2{x^2} + x + 1 - 2{x^2} + 2x = 0\\
\to - 4{x^2} + 3x + 1 = 0\\
\to \left[ \begin{array}{l}
x = 1\left( l \right)\\
x = - \dfrac{1}{4}
\end{array} \right.
\end{array}\)
