Bài 1:
a) Ta có: \(4\left(x-5\right)+2x=-13\)
\(\Leftrightarrow6x-20=-13\)
\(\Leftrightarrow6x=7\)
hay \(x=\dfrac{7}{6}\)
b) Ta có: \(\dfrac{2x-1}{4}-3=\dfrac{1+x}{6}\)
\(\Leftrightarrow\dfrac{3\left(2x-1\right)}{12}-\dfrac{36}{12}=\dfrac{2\left(x+1\right)}{12}\)
Suy ra: \(6x-3-36=2x+2\)
\(\Leftrightarrow4x=41\)
hay \(x=\dfrac{41}{4}\)
c) ĐKXĐ: \(x\notin\left\{0;3\right\}\)
Ta có: \(\dfrac{x+3}{x-3}-\dfrac{8}{x}=\dfrac{18}{x^2-3x}\)
\(\Leftrightarrow\dfrac{x\left(x+3\right)}{x\left(x-3\right)}-\dfrac{8\left(x-3\right)}{x\left(x-3\right)}=\dfrac{18}{x\left(x-3\right)}\)
Suy ra: \(x^2+3x-8x+24-18=0\)
\(\Leftrightarrow x^2-5x+6=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=3\left(loại\right)\end{matrix}\right.\)
Vậy: S={2}
Bài 2:
a) Ta có: 3x-11>5-x
\(\Leftrightarrow3x+x>5+11\)
\(\Leftrightarrow4x>16\)
hay x>4
Vậy: S={x|x>4}
b) Ta có: \(\dfrac{2x+1}{3}-\dfrac{3x-2}{6}\ge x+\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{2\left(2x+1\right)}{6}-\dfrac{3x-2}{6}\ge\dfrac{6x}{6}+\dfrac{1}{6}\)
Suy ra: \(4x+2-3x+2-6x-1\ge0\)
\(\Leftrightarrow-5x\ge-3\)
hay \(x< \dfrac{3}{5}\)
Vậy: S={x|\(x< \dfrac{3}{5}\)}
