2/a/ \(x\left(x+1\right)^2\)

b/ \(\left(y+x\right)\left(y-1\right)\)

c/ \(\left(x+1\right)\left(3x-10\right)\)

d/ \(-3\left(2z+x-y\right)\left(2z+y-x\right)\)

21(x-y)²-7(y-x)³ 

=21(x²-y²)-7(y²-x²)

=21x²-21y²-7y²+7x²

=28x²-28y²

=28(x²-y²)

Cưu là mình vs (x^2+x)^2-2(x^2+x)-15

      \(27x^3-27x^2+18x-4\)

\(=27x^3-9x^2-18x^2+6x+12x-4\)

\(=9x\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)\)

\(=\left(3x-1\right)\left(9x^2-6x+4\right)\)

       \(x^2+2xy+y^2-x-y-12\)

\(=\left(x+y\right)^2-\left(x+y\right)-12\)

\(=\left(x+y\right)^2-4\left(x+y\right)+3\left(x+y\right)-12\)

\(=\left(x+y\right)\left(x+y-4\right)+3\left(x+y-4\right)=\left(x+y+3\right)\left(x+y-4\right)\)  \(P=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

   \(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\) (nhóm 2 cái đầu với cuối lại với nhau, 2 cái giữa vào 1 nhóm)

Đặt \(x^2+7x+11=a\)

Ta có: \(P=\left(a-1\right)\left(a+1\right)-24\)

\(=a^2-25=\left(a-5\right)\left(a+5\right)\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

d,   \(4x^4-32x^2+1\)

\(=4x^4+4x^2+1-36x^2\)

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

1) \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12=x^4+x^3+2x^2+x^3+x^2+2x+x^2+x+2-12\)

\(=x^4+2x^3+4x^2+3x-10=\left(x^4+2x^3\right)+\left(4x^2+8x\right)+\left(-5x-10\right)\)

\(=x^3.\left(x+2\right)+4x.\left(x+2\right)-5.\left(x+2\right)=\left(x+2\right)\left(x^3+4x-5\right)\)

\(=\left(x+2\right)\left(x^3-x^2+x^2-x+5x-5\right)=\left(x+2\right)\left(x-1\right)\left(x^2+x+5\right)\)

2) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left[\left(x+2\right)\left(x+5\right)\right].\left[\left(x+3\right)\left(x+4\right)\right]-24\)

\(=\left(x^2+7x+10\right).\left(x^2+7x+12\right)-24\)

Đặt  \(a=x^2+7x+10\) thì ta có :\(a.\left(a+2\right)-24=a^2+2a-24=\left(a^2+2a+1\right)-25=\left(a+1\right)^2-5^2\)

\(=\left(a+1+5\right)\left(a+1-5\right)=\left(a+6\right)\left(a-4\right)\)

Thay a , ta có :

\(\left(x^2+7x+10+6\right)\left(x^2+7x+10-4\right)=\left(x^2+7x+16\right).\left(x^2+x+6x+6\right)\)

\(=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)

B1 :

a) (2x - 1)²

Bài 1: 

\(=3x^3y-6x^2y^2+15xy\)

Bài 2: 

\(=\left(x+y\right)^2-25=\left(x+y+5\right)\left(x+y-5\right)\)

\(x^2+2xy-25+y^2\\ =\left(x^2+2xy+y^2\right)-5^2\\ =\left(x+y\right)^2-5^2\\ =\left(x+y-5\right)\left(x+y+5\right)\)