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Ta có
a, x2-x-y2-y
=x2-y2-(x+y)
=(x-y)(x+y) - (x+y)
=(x+y)(x-y-1)
b, x2-2xy+y2-z2
=(x-y)2-z2
=(x-y-z)(x-y+z)
a) 16x2-(x2+4)2= (4x)2-(x2+4)2
= (4x-x2-4)(4x+x2+4)
\(\text{b) 27x^3-54x^2+36x-8=[(3x)^3-3.(3x)^2.2+3.3x.2^2-2^3}]\)
= (3x-2)3
\(\text{c) (x+y)^3 - (x-y)^3= (x+y-x+y)[(x+y)^2+(x+y)(x-y)+(x-y)^2]}\)
=2y(x2+2xy+y2+x2-y2+x2-2xy+y2)
= 2y(3x2+y2)
a) \(-y^2+\dfrac{1}{9}\)
\(=-\left(y^2-\left(\dfrac{1}{3}\right)^2\right)\)
\(=-\left(y+\dfrac{1}{3}\right)\left(y-\dfrac{1}{3}\right)\)
b) \(4^4-256\)
\(=4^4-4^4\)
\(=0\)
a/ \(=2xy+6x^2-y^2-3xy=2x\left(y+3x\right)-y\left(y+3x\right)=\left(2x-y\right)\left(y+3x\right)\)
b/ \(=2x^2+5xy+y^2\)
c/ \(=-4y^2+6xy-4xy+6x^2=-2y\left(2y-3x\right)-2x\left(2y-3x\right)=\left(-2y-2x\right)\left(2y-3x\right)=-2\left(y+x\right)\left(2y-3x\right)\)
d/ \(=-4y^2-2xy+4xy+2x^2=-2y\left(2y+x\right)+2x\left(2y+x\right)=\left(-2y+2x\right)\left(2y+x\right)=-2\left(y-x\right)\left(2y+x\right)\)
a/ \(=2xy+x^2-2y^2-xy=x\left(2y+x\right)-y\left(2y+x\right)=\left(x-y\right)\left(2y+x\right)\)
b/ \(=x^2-3xy-2y^2\)
c/ \(=2xy-6y^2+x^2-3xy=2y\left(x-3y\right)+x\left(x-3y\right)=\left(2y+x\right)\left(x-3y\right)\)
d/ \(=xy-2y^2+2x^2-4xy=y\left(x-2y\right)+2x\left(x-2y\right)=\left(y+2x\right)\left(x-2y\right)\)
a)(x-y)(x+2y)
b) nghiệm lẻ>mìh nghĩ là sai đề
c)(x-3y)(x+2y)
d)(x-2y)(2x+y)
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li-ke cho mk nha bnLe Ngoc Bao Chau
x^2 + xy - 2y^2 = ( x^2 - 2xy + y^2 ) + 3xy - 3y^2 = ( x - y )^2 + 3y.( x - y ) = ( x - y )(x + 2y)