\(x^3-4x^2-xy^2+4x=x\left(x^2-4x+4-y^2\right)=x\left[\left(x-2\right)^2-y^2\right]\)
\(=x\left(x-2-y\right)\left(x-2+y\right)\)
b.
\(x^3+x^2-x-1=x^2\left(x+1\right)-\left(x+1\right)=\left(x+1\right)\left(x^2-1\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(x+1\right)=\left(x+1\right)^2\left(x-1\right)\)
c.
\(16x-5x^2-3=15x-5x^2+x-3=5x\left(3-x\right)-\left(3-x\right)=\left(3-x\right)\left(5x-1\right)\)
a) Xem lại đề
b) \(x^3+x^2-x-1=x^3-1+x^2-x=\left(x-1\right)\left(x^2+x+1\right)+x\left(x-1\right)=\left(x-1\right)\left(x^2+x+1+x\right)=\left(x-1\right)\left(x^2+2x+1\right)=\left(x-1\right)\left(x+1\right)^2\)
c)\(16x-5x^2-3=-5x^2+15x+x-3=-5x\left(x-3\right)+\left(x-3\right)=\left(x-3\right)\left(-5x+1\right)\)
c) 2x2(x +1) + 4x(x +1); d) 2 x(y - 1) - 2