Bài 1:
\(a,=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\\ b,=4x^4+4x^2y^2+y^4-9y^4=\left(2x^2+y^2\right)^2-9y^4\\ =\left(2x^2+y^2-3y^2\right)\left(2x^2+y^2+3y^2\right)\\ =\left(2x^2-2y^2\right)\left(2x^2+4y^2\right)\\ =4\left(x-y\right)\left(x+y\right)\left(x^2+2y^2\right)\)
Bài 2:
\(a,=27\left(x^3-\dfrac{1}{729}\right)=27\left(x-\dfrac{1}{9}\right)\left(x^2+\dfrac{1}{9}x+\dfrac{1}{81}\right)\\ b,=3x\left(x-y\right)-5\left(x-y\right)=\left(3x-5\right)\left(x-y\right)\\ c,=x^3+x^2-x^2-x-6x-6\\ =\left(x+1\right)\left(x^2-x-6\right)\\ =\left(x+1\right)\left(x^2-3x+2x-6\right)\\ =\left(x+1\right)\left(x-3\right)\left(x+2\right)\)
Bài 3:
\(a,=\left(ay^2-by^2\right)-\left(4ay-4by\right)+\left(4a-4b\right)\\ =\left(a-b\right)\left(y^2-4y+4\right)=\left(a-b\right)\left(y-2\right)^2\\ b,=3\left(x^2+16+8x-4y^2\right)\\ =3\left[\left(x+4\right)^2-4y^2\right]\\ =3\left(x+2y+4\right)\left(x-2y+4\right)\)
Bài 1:
a,=3(x+y)−(x+y)2=(x+y)(3−x−y)b,=4x4+4x2y2+y4−9y4=(2x2+y2)2−9y4=(2x2+y2−3y2)(2x2+y2+3y2)=(2x2−2y2)(2x2+4y2)=4(x−y)(x+y)(x2+2y2)a,=3(x+y)−(x+y)2=(x+y)(3−x−y)b,=4x4+4x2y2+y4−9y4=(2x2+y2)2−9y4=(2x2+y2−3y2)(2x2+y2+3y2)=(2x2−2y2)(2x2+4y2)=4(x−y)(x+y)(x2+2y2)
Bài 2:
