\(\dfrac{6x-2}{x+4}=5\Leftrightarrow6x-2=5x+20\)
\(\Leftrightarrow x=22\)
\(\dfrac{6x-2}{x+4}=5\)
ĐKXĐ: \(x\ne-4\)
\(\dfrac{6x-2}{x+4}=\dfrac{5\left(x+4\right)}{x+4}\)
\(6x-2=5x+20\)
\(6x-5x=20+2\)
\(x=22\)
Vậy \(S=\left\{22\right\}\)
\(\dfrac{6x-2}{x+4}=5\left(dkxd:x\ne-4\right)\)
Suy ra :
\(6x-2=5\left(x+4\right)\)
\(\Rightarrow6x-2=5x+20\)
\(\Rightarrow x=22\left(tmdk\right)\)
Vậy \(S=\left\{22\right\}\)
\(\dfrac{6x-2}{x+4}=5\text{ĐKXĐ:}x\ne-4\)
\(\Leftrightarrow\dfrac{6x-2}{x+4}=\dfrac{5\left(x+4\right)}{x+4}MTC:x+4\)
\(\Rightarrow6x-2=5x+20\)
\(\Leftrightarrow6x-2-5x-20=0\)
\(\Leftrightarrow x-22=0\)
\(\Leftrightarrow x=22\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{22\right\}\)
