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

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

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

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

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

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

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

\(=\left(x^2+4x+8\right)\left(x^2+6x+8\right)+x\left(x^2+6x+8\right)\)

\(=\left(x^2+4x+8+x\right)\left(x^2+6x+8\right)=\left(x^2+5x+8\right)\left(x^2+6x+8\right)\)

Ta có: \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)

\(=\left(x^2+5x+8\right)\left(x^2+6x+8\right)\)

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

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

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

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

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

bài 2 là tìm X nha mn

\(1,\\ a,=x\left(2x+3y-5\right)\\ b,=x\left(x-2y\right)+\left(x-2y\right)=\left(x+1\right)\left(x-2y\right)\\ 2,\\ a,\Leftrightarrow x\left(x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\\ b,\Leftrightarrow x\left(x-2y\right)+\left(x-2y\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x-2y\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2y\left(y\in R\right)\end{matrix}\right.\)

\(x^2-4x+3\)

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

\(=x\left(x-1\right)-3\left(x-1\right)\)

\(=\left(x-1\right)\left(x-3\right)\)

x²-4x+3

=x.x-x-3x-3

=x.(x-1)-3.(x-1)

=(x-1)(x-3)

x² - 4x + 3

= x² - x - 3x + 3 ( dùng phương pháp tách nhân tử )

= ( x² - x ) - ( 3x - 3 ) ( vì đằng trc có đấu " - " nên ta đổi đáu trong ngoặc )

= x( x - 1 ) - 3(x - 1 ) ( đặt nhân tử chung )

= ( x - 1 ) ( x - 3 )

hok tốt !

\(\left(x^2+x\right)^2+4x^2+4x-12=\left[\left(x^2+x\right)^2+4\left(x^2+x\right)+4\right]-16=\left(x^2+x+2\right)-4^2=\left(x^2+x+2-4\right)\left(x^2+x+2+4\right)=\left(x^2+x-2\right)\left(x^2+x+6\right)=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)

\(\left(x^2+x\right)^2+4x^2+4x-12\\ =\left(x^2+x+2\right)-4\\ =\left(x^2+x-2\right)\left(x^2+x+6\right)\)

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

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

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

(1 + x²)² - 4x(1 - x²)

= (1 + x²)(1 + x²) - 4x(1 - x²)

= (1 + x² - 4x)(1 + x² - 1 + x²)

= 2x²(x² - 4x + 1)

Ta có: \(\left(x^2+1\right)^2+4x\left(x^2-1\right)\)

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

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

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

\(\left(x^2-5x\right)^2-3x^2+15x-18\)

\(=\left(x^2-5x\right)^2-3\left(x^2-5x\right)-18\)

\(=\left(x^2-5x-6\right)\left(x^2-5x+3\right)\)

\(=\left(x^2-5x+3\right)\left(x-6\right)\left(x+1\right)\)

\(=\left(x^2-5x\right)^2-3\left(x^2-5x\right)-18\\ =\left(x^2-5x\right)^2-6\left(x^2-5x\right)+3\left(x^2-5x\right)-18\\ =\left(x^2-5x\right)\left(x^2-5x-6\right)+3\left(x^2-5x-6\right)\\ =\left(x^2-5x+3\right)\left(x^2-5x-6\right)\\ =\left(x-6\right)\left(x+1\right)\left(x^2-5x+3\right)\)

\(=x^4-10x^3+25x^2-3x^2+15x-18=x^4-10x^3+22x^2+15x-18=x^4+x^3-11x^3-11x^2+33x^2+33x-18x-18=x^3\left(x+1\right)-11x^2\left(x+1\right)+33x\left(x+1\right)-18\left(x+1\right)=\left(x+1\right)\left(x^3-11x^2+33x-18\right)=\left(x+1\right)\left(x^3-6x^2-5x^2+30x+3x-18\right)=\left(x+1\right)\left[x^2\left(x-6\right)-5x\left(x-6\right)+3\left(x-6\right)\right]=\left(x+1\right)\left(x-6\right)\left(x^2-5x\right)=\left(x+1\right)\left(x-6\right)x\left(x-5\right)\)

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

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

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

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

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

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

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

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

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