Question

hộ mik vs 

Answer 1

Bài 3:

\(\left(x+2\right)^3+\left(x-2\right)^3+x^3-3x\left(x+2\right)\left(x-2\right)\\ =\left(x^3+6x^2+12x+8\right)+\left(x^3-6x^2+12x-8\right)+x^3-3x\left(x^2-4\right)\\ =x^3+6x^2+12x+8+x^3-6x^2+12x-8-3x^3+12x\\ =\left(x^3+x^3+x^3-3x^3\right)+\left(6x^2-6x^2\right)+\left(12x+12x+12x\right)+\left(8-8\right)\\ =36x\)  

Bài 4:

Ta có: 

\(VT=\left(x+y\right)^3-\left(x-y\right)^3\\ =\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\\ =\left(x+y-x+y\right)\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\\ =2y\left(3x^2+y^2\right)=VP\)