\(\Leftrightarrow5\left(x-4\right)+2x=2\left(x+1\right)\)
=>5x-20+2x=2x+2
=>7x-20=2x+2
=>5x=22
hay x=22/5
\(ĐK:x\ne-1;4\)
\(\Rightarrow\dfrac{5\left(x-4\right)+2x}{\left(x+1\right)\left(x-4\right)}=\dfrac{2\left(x+1\right)}{\left(x+1\right)\left(x-4\right)}\)
\(\Leftrightarrow5\left(x-4\right)+2x=2\left(x+1\right)\)
\(\Leftrightarrow5x-20+2x=2x+2\)
\(\Leftrightarrow5x=22\)
\(\Leftrightarrow x=\dfrac{22}{5}\left(tm\right)\)
Vậy \(S=\left\{\dfrac{22}{5}\right\}\)
ĐKXD: X ≠1;4
5(x-4)+2x=2(x+1)
5x-20+2x=2x+2
5x+2x-2x=20+2
5x=22
X=22/5 (22 phần 5)
