a) Ta có: \(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Vậy: x∈{-5;2}
b) Ta có: \(2\left(x-3\right)\left(x^2+1\right)+15x-5x^2=0\)
\(\Leftrightarrow2\left(x-3\right)\left(x^2+1\right)+5x\left(3-x\right)=0\)
\(\Leftrightarrow2\left(x-3\right)\left(x^2+1\right)-5x\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[2\left(x^2+1\right)-5x\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x^2-5x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x^2-4x-x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[2x\left(x-2\right)-\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{3;2;\frac{1}{2}\right\}\)
c) Ta có: \(\left(x+2\right)\left(3-4x\right)=x^2+4x+4\)
\(\Leftrightarrow\left(x+2\right)\left(3-4x\right)=\left(x+2\right)^2\)
\(\Leftrightarrow\left(x+2\right)\left(3-4x\right)-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)\left[\left(3-4x\right)-\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right)\left(3-4x-x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(1-5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\1-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\5x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\frac{1}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{-2;\frac{1}{5}\right\}\)
