12 Phân tích các đa thức sau thành nhân tử:
a) \(4x^3-8x^2+4xy^3\) ; b) \(x^2+2xy+y^2-36\) ; c) \(x^2-2xy+y^2-25\) ; d) \(x^2-5x+2xy-5y+y^2\)
e) \(49+2xy-x^2-y^2\) ; f) \(3x^2-6x+3-3y^2\) ; g) \(2x^3+4x^2+2x\) ; h) \(3x^2-6x+3-3y^2\)
i) \(x^3-2x^2y+xy^2-64x\) ; k) \(3x+3y-x^2-2xy-y^2\)
Thank bạn @Phùng Khánh Linh nhé
\(a.4x^3-8x^2+4xy^3=4x\left(x^2-8x+y^3\right)\)
\(b.x^2+2xy+y^2-36=\left(x+y\right)^2-36=\left(x+y-6\right)\left(x+y+6\right)\) \(c.x^2-2xy+y^2-25=\left(x-y\right)^2-25=\left(x-y-5\right)\left(x-y+5\right)\) \(d.x^2-5x+2xy-5y+y^2=\left(x+y\right)^2-5\left(x+y\right)=\left(x+y\right)\left(x+y-5\right)\) \(e.49+2xy-x^2-y^2=-\left(x^2-2xy+y^2-49\right)=-\left[\left(x-y\right)^2-49\right]=-\left(x-y-7\right)\left(x-y+7\right)\) \(f.3x^2-6x+3-3y^2=3\left(x^2-2x-y^2+1\right)\)
\(g.2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)\left(x+1\right)\)
\(h,\) giống câu f.
\(i.x^3-2x^2y+xy^2-64x=x\left(x^2-2xy+y^2-64\right)=x\left[\left(x-y\right)^2-64\right]=x\left(x-y-8\right)\left(x-y+8\right)\) \(k.3x+3y-x^2-2xy-y^2=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\)