Câu 5:
a) Ta có: \(\dfrac{5x-2}{3}+x=1+\dfrac{5-3x}{2}\)
\(\Leftrightarrow\dfrac{2\left(5x-2\right)}{6}+\dfrac{6x}{6}=\dfrac{6}{6}+\dfrac{3\left(5-3x\right)}{6}\)
\(\Leftrightarrow10x-4+6x=6+15-9x\)
\(\Leftrightarrow16x-4+9x-21=0\)
\(\Leftrightarrow25x=25\)
hay x=1
Vậy: S={1}
b) ĐKXĐ: \(x\notin\left\{3;-1\right\}\)
Ta có: \(\dfrac{2}{2x-6}+\dfrac{2}{2x+2}+\dfrac{2x}{\left(x+1\right)\left(3-x\right)}=0\)
\(\Leftrightarrow\dfrac{1}{x-3}+\dfrac{1}{x+1}-\dfrac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\dfrac{x+1+x-3-2x}{\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow-2=0\)(Vô lý)
Vậy: \(S=\varnothing\)
Câu 6:
a) Ta có: \(-5x\le x+2\)
nên \(-5x-x\le2\)
\(\Leftrightarrow-6x\le2\)
hay \(x\ge-\dfrac{1}{3}\)
Vậy: S={x|\(x\ge-\dfrac{1}{3}\)}
b) Ta có: \(\dfrac{5x+3}{4}-\dfrac{9x+2}{5}< \dfrac{7-3x}{8}\)
\(\Leftrightarrow\dfrac{10\left(5x+3\right)}{40}-\dfrac{8\left(9x+2\right)}{40}< \dfrac{5\left(7-3x\right)}{40}\)
\(\Leftrightarrow50x+30-72x-16< 35-15x\)
\(\Leftrightarrow-7x< 21\)
hay x>-3
Vậy: S={x|x>-3}
