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

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

\(9x^2+6x-4y^2+4y\)

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

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

\(=2\left(x^2-y^2\right)-6\left(x+y\right)=2\left(x-y\right)\left(x+y\right)-6\left(x+y\right)=\left(x+y\right)\left(2x-2y-6\right)\)                                                                                    Đảm bảo chuẩn ko cần chỉnh (•••

  check mk nhá
 

2X²-2Y²-6X-6Y

=2(X²-Y²) +6(X-Y)

=2(X-Y)(X+Y)+3.2(X-Y)

=2(X-Y)(X+Y+3X-3Y)

=2(X-Y)(4X-2Y)

=4(X-Y)(2X-Y)

\(2x^2-2y^2-6x-6y\)

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

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

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

4x+5xy-6x+9x²=9x²-2x+5xy=x(9x-2+5y)

hình như đề sai rồi

x²-y²+6x+6y = (x²-y²)+(6x+6y) = (x-y)(x+y)+6(x+y) = (x-y-6)(x+y)

x³ - 9x² + 7x + 1

= x³ - x2 - 8x² + 8x - x + 1

= (x³ - x²) - (8x² - 8x) - (x - 1)

= x²(x - 1) - 8x(x - 1) - (x - 1)

= (x - 1)(x² - 8x -1)

đề đúng ko

Đa thức này không phân tích được nhé bạn

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

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

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

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

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

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

\(4\left(x^2y^2+z^2t^2+2xyzt\right)-\left(x^2+y^2-z^2-t^2\right)^2\)

\(=\left[2\left(xy+zt\right)\right]^2-\left(x^2+y^2-z^2-t^2\right)^2\)

\(=\left(2xy+2zt\right)^2-\left(x^2+y^2-z^2-t^2\right)^2\)

\(=\left(2xy+2zt-x^2-y^2+z^2+t^2\right)\left(2xy+2zt+x^2+y^2-z^2-t^2\right)^2\)

Ta có: \(4\left(x^2y^2+2xyzt+z^2t^2\right)-\left(x^2+y^2-z^2-t^2\right)^2\)

\(=\left(2xy+2tz\right)^2-\left(x^2+y^2-z^2-t^2\right)^2\)

\(=\left(2xy+2tz-x^2-y^2+z^2+t^2\right)\left(2xy+2tz+x^2+y^2-z^2-t^2\right)\)

\(=\left[-\left(x^2-2xy+y^2\right)+\left(z^2+2tz+t^2\right)\right]\left[\left(x^2+2xy+y^2\right)-\left(t^2-2tz+z^2\right)\right]\)

\(=\left(z+t-x+y\right)\left(z+t+x-y\right)\left(x+y-t+z\right)\left(x+y+t-z\right)\)

\(=6x^2+9x+4x+6\)

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

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

\(\)

6x² + 13x + 6

= 2x(3x + 2) + 3(3x + 2)

= (3x + 2)(2x + 3)

6x²  +13x+6 = 6x²  + 4x +9x +6 = 2x(3x+2) + 3(3x+2) = (3x+2)(2x+3)