phân tích đa thứ sau thành nhân tử: 3x-3y+2x^2-2xy
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a) \(x^2-y^2-3x+3y\)
\(=\left(x-y\right)\left(x+y\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-3\right)\)
b) \(2x+2y-x^2+y^2\)
\(=2\left(x+y\right)-\left(x^2-y^2\right)\)
\(=2\left(x+y\right)-\left(x-y\right)\left(x+y\right)\)
\(=\left(x+y\right)\left(2-x+y\right)\)
c) \(x^2-16+y^2+2xy\)
\(=x^2+y^2+2xy-16\)
\(=\left(x+y\right)^2-16\)
\(=\left(x+y+4\right)\left(x+y-4\right)\)
a) \(x^2-y^2-3x+3y\)
\(=\left(ax+y\right)\left(ax-y\right)-3.\left(x-y\right)\)
b) \(2x+2y-x^2+y^2\)
\(=2\left(x+y\right)-\left(x+y\right)\left(x-y\right)\)
c) \(x^2-16+y^2+2xy\)
\(=\left(x+y\right)\left(x-y\right)+2xy-16\)
\(a,2x^3y-2xy=2xy\left(x^2-1\right)=2xy\left(x-1\right)\left(x+1\right)\)
\(b,x^2-2x-4x^2-4x=-3x^2-2x-4x\\ =-3x^2-6x=-3\left(x^2+2x\right)=-3x\left(x+2\right)\)
a) x3-2x2-x+2
=x(x2-1)+2(-x2+1)
=x(x2-1)-2(x2-1)
=(x2-1)(x-2)
b)
x2+6x-y2+9
=x2+6x+9-y2
=(x+3)2-y2
=(x+3-y)(x+3+y)
2x(x-2)+2y(x-2)= (x-2)(2x+2y)=2(x-2)(x+y)
b,2(xy+xyz-2x-2z)
c, 3(x^2-xy-x-y)
a) Ta có : 2x2 - 4x + 2xy - 4y
= 2x(x - 2) + 2y(x - 2)
= (x - 2)(2x + 2y)
= 2(x - 2)(x + y)
\(-y^2+2xy-x^2+3x-3y\)
\(=-\left(x^2-2xy+y^2\right)+3\left(x-y\right)\)
\(=-\left(x-y\right)^2+3\left(x-y\right)\)
\(=\left(x-y\right)\left(-x+y+3\right)\)
a) xy+3x-7y-21
=x(y+3)-7(x+3)
=(x-7)(y+3)
b)2xy-15-6x-5y
=2x(y-3)-5(-3+y)
=(2x-5)(y-3)
c)2x^2y+2xy^2-2x-2y
=2x(xy-1)+2y(xy-1)
=(2x+2y)(xy-1)
x(x+3)-5x(x-5)-5(x+3)
=(x-5)(x+3)-5x(x-5)
=(x-5)(x+3-5x)
Câu cuối mình bị nhầm dòng cuối phải là (x-5)(x+3+x-5)=(x-5)(2x-2)nha bạn
\(a,3x^2-6x+9x^2=12x^2-6x=6x\left(2x-1\right)\\ b,3x^2+5y-3xy-5x=\left(3x^2-3xy\right)-\left(5x-5y\right)=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\\ c,3y^2-3z^2+3x^2+6xyz=3\left(y^2-z^2+x^2+2xyz\right)\\ d,x^2-25-2xy+y^2=\left(x-y\right)^2-5^2=\left(x-y-5\right)\left(x-y+5\right)\)
\(3x-3y+2x^2-2xy\)
\(=3\left(x-y\right)+2x\left(x-y\right)\)
\(=\left(x-y\right)\left(3+2x\right)\)