a, \(5x-15y=5\left(x-3y\right)\)

b, \(12y\left(2x-5y\right)+6xy\left(5-2x\right)=12y\left(2x-5\right)-6xy\left(2x-5\right)\)

\(=6y\left(2-x\right)\left(2x-5\right)\)

c, \(x^2-7x+12=x^2-3x-4x+12=\left(x-4\right)\left(x-3\right)\)

\(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)=0\)

\(\Leftrightarrow x^5-3x^4+3x^2-1-\left(x^6-x^4+x^4-x^2+x^2-1\right)=0\)

\(\Leftrightarrow x^5-3x^4+3x^2-x^6=0\)

\(\Leftrightarrow x^5\left(1-x\right)-3x^2\left(x^2-1\right)=0\Leftrightarrow x^5\left(1-x\right)-3x^2\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow-x^5\left(x-1\right)-3x^2\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[-x^5-3x^3-3x^2\ne0\right]=0\Leftrightarrow x=1\)

\(a^3-8a^2+16a=a\left(a^2-2.4a+4^2\right)=a\left(a-4\right)^2\)

\(x^3+2x^2-4x-8=x^2\left(x+2\right)-4\left(x+2\right)=\left(x^2-4\right)\left(x+2\right)=\left(x-2\right)\left(x+2\right)^2\)

a, Ta có : \(M=x^2-x=0\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow x=0;1\)

b, Ta có : \(M=N\) hay \(x^2-x=\left(x-1\right)^3-x^2\left(x-3\right)-2\)

\(x^2-x=x^3-3x^2+3x-1-x^3+3x^2-2\)

\(x^2-x=3x-3\Leftrightarrow x^2-4x+3=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\Leftrightarrow x=3;1\)

\(x^3+x-3x^2-3=0\Leftrightarrow x\left(x^2+1\right)-3\left(x^2+1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2+1\ne0\right)=0\Leftrightarrow x=3\)

\(x^3-3.x^2+x-3=0\)

\(\Rightarrow\)\(x^2.\left(x-3\right)+\left(x-3\right)=0\)

                  \(\left(x^2+1\right).\left(x-3\right)=0\)\(\Rightarrow\orbr{\begin{cases}x^2+1\\x-3\end{cases}}=0\)

Với :  \(x^2+1=0\Rightarrow x=\varnothing\)nhưng giá trị này làm cho biểu thức không có nghĩa, loại

         \(x-3=0\Rightarrow x=3\)

Vậy  \(x=3\)