\(2,=x^2-3^2=\left(x-3\right)\left(x+3\right)\\ 3,=\left(x+y-x+y\right)\left(x+y+x-y\right)\\ =2y\cdot2x=4xy\)
x^2-9=x^2-3^2=(x+3)(x-3)
(x+y)^2-(x-y)^2=(x+y+x-y)(x+y-x-y)=2x
x2-32=(x-3)(x+3)
(x+y)2-(x-y)2=(x+y-x+y)(x+y+x-y)=2y⋅2x=4xy
Câu 2:
Ta có:
\(x^2-9=x^2-3^2=\left(x-3\right)\left(x+3\right)\)
Câu 3:
Ta có:
\(\left(x+y\right)^2-\left(x-y\right)^2=\left[\left(x+y\right)+\left(x-y\right)\right]\left[\left(x+y\right)-\left(x-y\right)\right]\)
\(=\left(x+y+x-y\right)\left(x+y-x+y\right)\)
\(=2x.2y=4xy\)