a)\(2x+3\left(x+4\right)-14=8\)
\(2x+3x+12-14=8\)
\(2x+3x-2=8\)
\(5x-2=8\)
\(5x=8+2\)
\(5x=10\)
\(x=10:5\)
\(x=2\)
b)\(x+2x+3x=4x+16\)
\(x\left(1+2+3\right)=4x+16\)
\(6x-4x=16\)
\(2x=16\)
\(x=16:2\)
\(x=8\)
c)\(2\left(x+1\right)+3\left(x+5\right)=0\)
\(2x+2+3x+15=0\)
\(5x+17=0\)
\(5x=0-17\)
\(5x=-17\)
\(x=-17:5\)
\(x=-\frac{17}{5}\)
d)\(\left(x-3\right).\left(x-5\right)=0\)
\(\hept{\begin{cases}x-3=0\\x-5=0\end{cases}\Rightarrow\hept{\begin{cases}x=3\\x=5\end{cases}}}\)
Vậy \(x\in\left\{3;5\right\}\)