Bài 2:
a) \(\left(3x-1\right)^2-16=\left(3x-5\right)\left(3x+3\right)\)
b) \(\left(5x-4\right)^2-49x^2=\left(-2x-4\right)\left(12x-4\right)\)
c) \(\left(2x+5\right)^2-\left(x-9\right)^2=\left(x+14\right)\left(3x-4\right)\)
d) \(\left(3x+1\right)^2-4\left(x-2\right)^2=\left(x+5\right)\left(5x-3\right)\)
e) \(9\left(2x+3\right)^2-4\left(x+1\right)=\left(2x+5\right)\left(10x+13\right)\)
f) ko rõ đề
g) \(\left(ax+by\right)^2-\left(ay+bx\right)^2=\left(ax+by-ay-bx\right)\left(ax+by+ay+bx\right)=\left[\left(ax-ay\right)+\left(by-bx\right)\right]\left[\left(ax+ay\right)+\left(by+bx\right)\right]=\left[a\left(x-y\right)-b\left(x-y\right)\right]\left[a\left(x+y\right)+b\left(x+y\right)\right]=\left(a-b\right)\left(x-y\right)\left(a+b\right)\left(x+y\right)=\left(a^2-b^2\right)\left(x^2-y^2\right)\)
h) \(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)=\left[\left(a^2-2ab+b^2\right)-9\right]\left[\left(a^2+2ab+b^2\right)-1\right]=\left[\left(a-b\right)^2-3^2\right]\left[\left(a+b\right)^2-1^2\right]=\left(a-b-3\right)\left(a-b+3\right)\left(a+b-1\right)\left(a+b+1\right)\)
i) \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2=\left(4x^2-3x-18-4x^2-3x\right)\left(4x^2-3x-18+4x^2+3x\right)=\left(-6x-18\right)\left(8x^2-18\right)=-6\left(x+3\right).2\left(4x^2-9\right)=-12\left(x+3\right)\left(2x-3\right)\left(2x+3\right)\)
