Phân tích đa thức thành nhân tử:
\(2x^2-3ax-9a^2\)
\(2x^2-17xy-9y^2\)
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1) \(x\left(4x+1\right)\)
2) \(3\left(x-3y\right)\)
3) \(\left(2x+1\right)\left(2x+1+2\right)=\left(2x+1\right)\left(2x+3\right)\)
2) \(x^4-5x^2+4\)
\(=x^4-x^2-4x^2+4\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2-4\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
a: =(2x-3y)^2-4(2x-3y)
=(2x-3y)(2x-3y-4)
b: =3x^2+21x-x-7
=(x+7)(3x-1)
c: =(3x-1)^4+2(3x-1)^2+1
=[(3x-1)^2+1]^2
d: =2x^3-2x^2-x^2+x+x-1
=(x-1)(2x^2-x+1)
a,\(x^2-y^2+2x+1=\left(x^2-2x+1\right)-y^2=\left(x+1\right)^2-y^2=\left(x+1-y\right)\left(x+1+y\right)\)b
\(4x^2-17xy+13y^2=\left(4x^2-4xy\right)+\left(13y^2-13xy\right)=4x\left(x-y\right)-13y\left(x-y\right)\)\(=\left(4x-13y\right)\left(x-y\right)\)
\(=4x^2-4xy-13xy+13y^2=4x\left(x-y\right)-13y\left(x-y\right)=\left(4x-13y\right)\left(x-y\right)\)
a) Ta có: \(x^2-2xy+y^2-2x+2y\)
\(=\left(x-y\right)^2-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-2\right)\)
b) Ta có: \(x^2-4x+4-x^2y+2xy\)
\(=\left(x-2\right)^2-xy\left(x-2\right)\)
\(=\left(x-2\right)\left(x-2-xy\right)\)
\(2x^2-3ax-9a^2\)
\(=2x^2-6ax+3ax-9a^2\)
\(=2x\left(x-3a\right)+3\left(x-3a\right)\)
\(=\left(x-3a\right)\left(2x+a\right)\)
\(2x^2-17xy-9y^2\)
\(=2x^2+xy-18xy-9y^2\)
\(=2x\left(2x+y\right)-9y\left(2x+y\right)\)
\(=\left(2x+y\right)\left(2x-9y\right)\)