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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)
#)Giải :
\(x^2-y^2+3x-3y=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\)
\(x^2-y^2+3x-3y\)
\(=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)\)
\(=\left(x-y\right)\left(x+x+3\right)\)
~ Rất vui vì giúp đc bn ~
\(3x+3y-x^2-xy\)
\(=\left(3x+3y\right)-\left(x^2+xy\right)\)
\(=3\left(x+y\right)-x\left(x+y\right)\)
\(=\left(3-x\right)\left(x+y\right)\)
\(-y^2+2xy-x^2+3x-3y\)
\(=-\left(x^2-2xy+y^2\right)+3\left(x-y\right)\)
\(=-\left(x-y\right)^2+3\left(x-y\right)\)
\(=\left(x-y\right)\left(-x+y+3\right)\)
Ta có: \(x^3y^3+x^2y^2+4=x^3y^3+8+x^2y^2-4=\left(xy+2\right)\left(x^2y^2-2xy+4\right)+\left(xy+2\right)\left(xy-2\right)\)
\(=\left(xy+2\right)\left(x^2y^2-xy+2\right)\)
2x^2-5xy-3y^2
= 2^x + xy - 6xy - 3y^2
= x(2x + y) - 3y(2x + y)
= (2x + y)(x - 3y)
Ta có: \(\left(2x+3y\right)^2-2\left(2x+3y\right)\)
\(=\left(2x+3y\right)\left(2x+3y-2\right)\)
\(x^3y-xy^3-2xy^2-xy\)
\(=xy\left(x^2-y^2-2y-1\right)\)
\(=xy\left[x^2-\left(y^2+2y+1\right)\right]\)
\(=xy\left[x^2-\left(y+1\right)^2\right]\)
\(=xy\left(x-y-1\right)\left(x+y+1\right)\)
\(x^2-3y^2=\left(x-\sqrt{3}y\right)\left(x+\sqrt{3}y\right)\)