Bài `2:`
`a)(x+2)^2-(x-3)(x+1)`
`=x^2+4x+4-x^2-x+3x+3`
`=6x+7`
`b)(x^3-2x^2+5x-10):(x-2)`
`=[x^2(x-2)+5(x-2)]:(x-2)`
`=(x-2)(x^2+5):(x-2)`
`=x^2+5`
bài 1;
a)\(6x^2-3xy\)
=\(3x.\left(2x-y\right)\)
b)\(x^2-y^2-6x+9\)
=\(\left(x^2-6x+9\right)-y^2\)
=\(\left(x-3\right)^2-y^2\)
=\(\left(x-3-y\right).\left(x-3+y\right)\)
c)\(x^2+5x-6\)
= \(\left(x-1\right).\left(x+6\right)\)
Bài 2
a)\(\left(x+2\right)^2-\left(x-3\right).\left(x+1\right)\)
=\(x^2-4x+4-x^2+x-3x-3\)
=\(-6x+1\)
b) ko bt
