BXD: \(\begin{array}{|c|cc|}\hline x&-\infty&&-2&&7&&+\infty\\\hline x+2&&-&0&+&|&+&\\\hline 7-x&&+&|&+&0&-\\\hline\end{array}\)
Xét khoảng \(x<-2\)
\(→-(x+2)+(7-x)=3x+4\\↔-x-2+7-x-3x-4=0\\↔-5x=-1\\↔x=\dfrac{1}{5}(KTM)\)
Xét khoảng \(-2\le x\le 7\)
\(→(x+2)+(7-x)=3x+4\\↔x+2+7-x=3x+4\\↔9-4=3x\\↔5=3x\\↔\dfrac{5}{3}=x(TM)\)
Xét khoảng \(x>7\)
\(→(x+2)-(7-x)=3x+4\\↔x+2-7+x=3x+4\\↔2x-5=3x+4\\↔x=-9(KTM)\)
Suy ra \(x=\dfrac{5}{3}\)
Vậy pt có tập nghiệm \(S=\{\dfrac{5}{3}\}\)
