a. \(x^3+3x^2-4=x^3+2x^2+x^2-4=x^2\left(x+2\right)+\left(x-2\right)\left(x+2\right)=\left(x+2\right)\left(x^2+x-2\right)\)
b. \(y^2+4y-12=y^2+4y+4-16=\left(y+2\right)^2-4^2=\left(y+2+4\right)\left(y+2-4\right)=\left(y+6\right)\left(y-2\right)\)
c. \(9x^2+6x-8=9x^2+6x+1-9=\left(3x+1\right)^2-3^2=\left(3x+1-3\right)\left(3x+1+3\right)=\left(3x-2\right)\left(3x+4\right)\)
d. \(2x^3-3x^2+3x-1=x^3-3x^2+3x-1+x^3=\left(x-1\right)^3+x^3=\left(x-1+x\right)\left[\left(x-1\right)^2-x\left(x-1\right)+x^2\right]=\left(2x-1\right)\left(x^2-2x+1-x^2+x+x^2\right)=\left(2x-1\right)\left(x^2-x+1\right)\)
e. \(\left(ax+by\right)^2-\left(ay+bx\right)^2=\left(ax+by+ay+bx\right)\left(ax+by-ay-bx\right)=\left[a\left(x+y\right)+b\left(x+y\right)\right]\left[a\left(x-y\right)+b\left(y-x\right)\right]=\left(x+y\right)\left(a+b\right)\left(x-y\right)\left(a-b\right)=\left(x^2-y^2\right)\left(a^2-b^2\right)\)
b, y2+4y-12
= y2-2y+6y-12
= y(y-2)+6(y-2)
= (y-2)(y+6)
c, 9x2+6x-8
= 9x2-12x+6x-8
= 3x(3x-4)+2(3x-4)
= (3x-4)(3x+2)
Câu d : Sửa đề
\(x^3-3x^2+3x-1=\left(x-1\right)^3\)
Câu e :
\(\left(ax+by\right)^2-\left(ay+bx\right)^2=\left(ax+by-ay-bx\right)\left(ax+by+ay+bx\right)\)
a)\(x^3+3x^2-4=x^3-x^2+4x^2-4x+4x-4\)
=\(x^2\left(x-1\right)+4x\left(x-1\right)+4\left(x-1\right)\)
=\(\left(x-1\right)\left(x+2\right)^2\)
b;c bạn kia làm r nhé
d)\(2x^3-3x^2+3x-1=2x^3-x^2-2x^2+x+2x-1\)
=\(x^2\left(2x-1\right)-x\left(2x-1\right)+\left(2x-1\right)=\left(2x-1\right)\left(x^2-x+1\right)\)
e)\(\left(ax+by\right)^2-\left(ay+bx\right)^2\)=\(\left(ax+by+bx+ay\right)\left(ax+by-ay-bx\right)=\left(a+b\right)\left(x+y\right)\left(a-b\right)\left(x-y\right)\)
