\(a,=x^2-9-x^2+6x-9=6x-18\\ b,=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x+y-5\right)\left(x-y\right)\)

a, <=>y²-3² <=> y² -9 (hằng đẳng thức số 3)

b, <=> m³+n³ ( hằng đẳng thức số 6)

c, <=> 2³-a³ (__________________số 7)

d, <=> (a-b-c-a+b-c )( a-b-c+a-b+c) 

<=> -2c*2a= -4ac

e, <=> (a-x-y-a-x+y) [(a-x-y) ²+(a-x-y)(a+x-y)+(a+x-y)²]

(Nhân phá ngoặc) -)

d <=> (1-x²)[(1+x²)²-x²)

<=> (1-x²)(1+2x²)

<=> 1+2x²-x²-2x⁴

<=> 1+x²-2x⁴

\(A=2^2+2^2+2^3+...+2^{2021}\\ 2A=2^3+2^3+2^4+...+2^{2022}\\ A=2^{2022}+2^3-2^2-2^2\)

\(2A=8+2^3+...+2^{2022}\)

\(\Leftrightarrow A=2^{2022}+8-4-2^2=2^{2022}\)

Ta có: \(A=\left(x-y-1\right)^3-\left(x-y+1\right)^3+6\left(x-y\right)^2\)

\(=\left(x-y-1-x+y-1\right)\left[\left(x-y-1\right)^2+\left(x-y-1\right)\left(x-y+1\right)+\left(x-y+1\right)^2\right]+6\left(x-y\right)^2\)

\(=-2\cdot\left[3\left(x-y\right)^2+1\right]+6\left(x-y\right)^2\)

\(=-6\left(x-y\right)^2+6\left(x-y\right)^2-2\)

=-2

\(A=\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)

\(=x^3-8-x^3-3x^2-3x-1+3x^2-3\)

\(=-3x-11\)

\(A,xy\left(2x^2-3\right)-x^2\left(5xy+y\right)+x^2y\\ =2x^3y-3xy-5x^3y-x^2y+x^2y\\ =\left(2x^3y-5x^3y\right)+\left(-x^2y+x^2y\right)-3xy\\ =-3x^3y-3xy\)

\(B,3xyz\left(y-2\right)-5yz\left(1-y\right)-8z\left(y^2-3\right)\\ =3xy^2z-6xyz-5yz+5y^2z-8y^2z+24z\\ =3xy^2z-6xyz+\left(5y^2z-8y^2z\right)-5yz+24z\\ =3xy^2z-6xyz-3y^2z-5yz+24z\)

Bài 1:

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