a) \(2x+5=\left(-10\right)-x\)
\(2x+x=-10-5\)
\(3x=-15\)
\(x=-15\div3\)
\(x=-5\)
\(5^2+125x=0\)
\(25+125x=0\)
\(125x=0-25\)
\(125x=-25\)
\(x=-25\div125\)
\(x=-\frac{1}{5}\)
\(7x^2-21x+5=5\)
\(7x^2-21x=5-5\)
\(7x^2-21x=0\)
\(7\left(x^2-3x\right)=0\)
\(7\left(x-3\right).x=0\)
\(\hept{\begin{cases}x-3=0\\x=0\end{cases}\Rightarrow\hept{\begin{cases}x=3\\x=0\end{cases}}}\)
Vậy \(x\in\left\{3;0\right\}\)
