\(x^6+y^6=\left(x^2\right)^3+\left(y^2\right)^3=\left(x^2+y^2\right).\left(x^4-x^2y^2+y^4\right)\\ ---\\ 0,04-9x^2=\left(0,2\right)^2-\left(3x\right)^2=\left(0,2-3x\right)\left(0,2+3x\right)\\ ---\\ 32x^2-2\left(y-1\right)^2=2\left[16x^2-\left(y-1\right)^2\right]=2\left[\left(4x\right)^2-\left(y-1\right)^2\right]\\ =2\left(4x-y+1\right)\left(4x+y-1\right)\)

Đúng hết mà?

\(\left(x^2+2x\right)^2+4\left(x^2+2x\right)+5\left(x^2+2x\right)+20\)

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

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

(x2+2x)+9x2+18x+20

=(x2+2x)+9(x2+2x)+20

Đặt t=x2+2x đc:

t+9t+20=10t+20=10(t+2)

Thay t=x2+2x vào đc:

10(x2+2x+2)

B

Ta có: 

\(6x^2+10xz+9x+15z\)

\(=\left(6x^2+10xz\right)+\left(9x+15z\right)\)

\(=2x.\left(3x+5z\right)+3.\left(3z+5z\right)\)

\(=\left(2x+3\right).\left(3x+5z\right)\)

vậy thừa số cần tìm là: \(3x+5z\)

Ta có:

6x²+10xz+9x+15z

=(6x²+9x)+(10xz+15z)

=3x(2x+3)+5z(2x+3)

=(2x+3)(3x+5z)

Anh(chị) xem lại đề nhé!!!

Em mới lớp 7 thôi nên có sai sót thì đừng k sai cho em nhé!!!

1: \(6x^2y-9xy^2+3xy\)

\(=3xy\left(2x-3y+1\right)\)

2: \(\left(4-x\right)^2-16\)

\(=\left(4-x-4\right)\left(4-x+4\right)\)

\(=-x\cdot\left(8-x\right)\)

3: \(x^3+9x^2-4x-36\)

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

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

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

2) \(\left(4-x\right)^2-16=\left(4-x\right)^2-4^2=\left(4-x-4\right)\left(4-x+4\right)=-x\left(8-x\right)\)

3) \(x^3+9x^2-4x-36\\ =\left(x^3-2x^2\right)+\left(11x^2-22x\right)+\left(18x-36\right)\\ =x^2\left(x-2\right)+11x\left(x-2\right)+18\left(x-2\right)\\ =\left(x^2+11x+18\right)\left(x-2\right)\\ =\left[\left(x^2+2x\right)+\left(9x+18\right)\right]\left(x-2\right)\\ =\left[x\left(x+2\right)+9\left(x+2\right)\right]\left(x-2\right)\\ =\left(x+2\right)\left(x+9\right)\left(x-2\right)\)

\(\left(a\right)\left(x^2+x\right)^2+9x^2+9x+14\)

\(\text{ Phân tích thành nhân tử}\)

\(\left(x^2+x+2\right)\left(x^2+x+7\right)\)

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

\(\text{ Phân tích thành nhân tử}\)

\(y^2+2xy-3y+x^2+2x-3\)

Xong rùi đấy !