Bài 1:
\(x^2+x-6=x^2+3x-2x+6\)
\(=x\left(x+3\right)-2\left(x+3\right)\)
\(=\left(x-2\right)\left(x+3\right)\)
\(b,x^4+2x^3+x^2=\left(x^2+x\right)^2\)
\(e,x^2+5x-6=x^2+6x-x-6\)
\(=x\left(x+6\right)-\left(x+6\right)=\left(x-1\right)\left(x+6\right)\)
\(f,5x^2+5xy-x-y=5x\left(x+y\right)-\left(x+y\right)=\left(5x-1\right)\left(x+y\right)\)\(g,7x-6x^2-2=-6x^2+3x+4x-2\)
\(=-3x\left(2x-1\right)+2\left(2x-1\right)=\left(2-3x\right)\left(2x-1\right)\)\(i,2x^2+3x-5=2x^2-2x+5x-5\)
\(=2x\left(x-1\right)+5\left(x-1\right)=\left(2x+5\right)\left(x-1\right)\)
\(j,16x-5x^2-3=-5x^2+15x+x-3\)
\(=-5x\left(x-3\right)+\left(x-3\right)=\left(5x-1\right)\left(x+3\right)\)
Bài 2,
\(a,5x\left(x-1\right)=x-1\)
\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(5x-1\right)\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5x-1=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=1\end{matrix}\right.\)
\(b,2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
