Em làm thử nếu sai thì thôi ạ (vì mới học lớp 6)

a) 

Ta có:

\(\left(a+b\right)^2-\left(a-b\right)^2=a^2.b^2-a^2:b^2\)

\(=a^2.b^2-a^2.\frac{1}{b^2}=a^2.\left(b^2-\frac{1}{b^2}\right)\)

Chắc thế ạ, em chỉ làm 1 phần vì sợ sai

a)(a+b)²-(a-b)²=(a+b+a-b)(a+b-a+b)=2a.2b=4ab

b)(a+b)³-(a-b)³-2ab³

=(a+b-a+b)[(a+b)²+(a+b)(a-b)+(a-b)²]-2ab³

=2a(a²+2ab+b²+a²-b²+a²-2ab+b²)-2ab³

=2a(3a²+b²)-2ab³

=6a³+2ab²-2ab³

c)(x+y+z)²-2(x+y+z)(x+y)+(x+y)²

=(x+y+z-x-y)²=z²

(a+b)^2+(a-b)^2

a)(x+y+z)² - 2(x+y+z)(x+y)+(x+y)²

=[(x+y+z)-(x-y)]²

=(x+y+z-x-y)²

=z²

b) (a+b)³ - (a - b)³ - 2b³

=[(a+b)-(a-b)][(a+b)²+(a+b)(a-b)+(a-b)²]-2b³

=(a+b-a+b)(a²+2ab+b²+a²-b²+a²-2ab+b²)-2b³

=2b(3a²+b²)-2b³

=6a²b+2b³-2b³

=6a²b

c) (a + b)² - (a - b)²=[a+b+(a-b)][a+b-(a-b)]=(a+b+a-b)(a+b-a+b)

                         =2a.2b=4ab

a) Ta có: (a+b)² - (a-b)²

        = (a+b+a-b)(a+b-a+b)

        =    2a.2b

        = 4ab

b) Ta có: (a+b)³ - (a-b)³ - 2b³

        = a³ + 3a²b + 3ab² + b³ - a³ + 3a²b - 3ab² + b³ - 2b³

        = 6a²b

c) Ta có: (x+y+z)² - 2(x+y+z)(x+y) + (x+y)²

        = (x+y+z-x-y)²

          = z²

Bài 1:

a) \(x\left(x-3\right)\left(x+3\right)-\left(x^2-2\right)\left(x^2+2\right)\)

\(=x\left(x^2-9\right)-x^4+4\)

\(=x^3-9x-x^4+4\)

\(=-x^4+x^3-9x+4\)

b) \(\left(y+2\right)\left(y-2\right)\left(y^2+4\right)-\left(y^2-3\right)\left(y^2+3\right)\)

\(=\left(y^2-4\right)\left(y^2+4\right)-y^4+9\)

\(=y^4-16-y^4+9\)

\(=-7\)

c) \(\left(a-b+c\right)^2+\left(b-c\right)^2+2\left(a-b+c\right)\left(c-b\right)\)

\(=\left(a-b+c\right)^2-2\left(a-b+c\right)\left(b-c\right)+\left(b-c\right)^2\)

\(=\left[\left(a-b+c\right)-\left(b+c\right)\right]^2\)

\(=\left(a-2b\right)^2\)

\(=a^2-4ab+4b^2\)