a) \(\left|3x-2\right|+5x=4x-10\)
\(\Leftrightarrow\left|3x-2\right|=-x-10\) \(\left(x\le-10\right)\)
\(\Leftrightarrow\left(3x-2\right)^2=\left(-x-10\right)^2\)
\(\Leftrightarrow8x^2-32x-96=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\) ( loại )
Vậy pt vô nghiệm.
b) \(3+\left|2x-5\right|>13\)
\(\Leftrightarrow\left|2x-5\right|>10\)
\(\Leftrightarrow\left(2x-5\right)^2>100\)
\(\Leftrightarrow4x^2-20x-75>0\)
\(\Leftrightarrow\left(2x-15\right)\left(2x+5\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}x>\frac{15}{2}\\x< \frac{-5}{2}\end{matrix}\right.\)
Vậy...
b) Ta có: 3+|2x+5|>13
⇒|2x+5|>10
\(\Rightarrow\left\{{}\begin{matrix}2x+5>3\\2x+5< -3\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}2x>-2\\2x< -8\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>-1\\x< -4\end{matrix}\right.\)
Vậy: -4<x<-1
\(\begin{array}{l} a)\left| {3x - 2} \right| + 5x = 4x - 10\\ \Leftrightarrow \left| {3x - 2} \right| + x = - 10\\ \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} 3x - 2 \ge 0\\ 3x - 2 + x = - 10 \end{array} \right.\\ \left\{ \begin{array}{l} 3x - 2 < 0\\ - \left( {3x - 2} \right) + x = - 10 \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x \ge \dfrac{2}{3}\\ 4x = - 8 \end{array} \right.\\ \left\{ \begin{array}{l} x < \dfrac{2}{3}\\ - 2x = - 12 \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x \ge \dfrac{2}{3}\\ x = - 2\left( {ktm} \right) \end{array} \right.\\ \left\{ \begin{array}{l} x < \dfrac{2}{3}\\ x = 6\left( {ktm} \right) \end{array} \right. \end{array} \right. \end{array}\)
Vậy phương trình vô nghiệm.
\(\begin{array}{l} b)3 + \left| {2x + 5} \right| > 13\\ \Leftrightarrow \left| {2x + 5} \right| > 10\\ \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} 2x + 5 \ge 0\\ 2x + 5 > 10 \end{array} \right.\\ \left\{ \begin{array}{l} 2x + 5 < 0\\ - \left( {2x + 5} \right) > 10 \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x \ge - \dfrac{5}{2}\\ x > \dfrac{5}{2}\left( {tm} \right) \end{array} \right.\\ \left\{ \begin{array}{l} x < - \dfrac{5}{2}\\ x < - \dfrac{{15}}{2}\left( {tm} \right) \end{array} \right. \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x > \dfrac{5}{2}\\ x < - \dfrac{{15}}{2} \end{array} \right. \end{array}\)
