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Đa thức này không phân tích được nhé bạn

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

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

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

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

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

bài 2 là tìm X nha mn

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

\(\left(x^2-2x-6\right)\left(x^2-2x-11\right)+6\)

\(=\left(x^2-2x\right)^2-17\left(x^2-2x\right)+66+6\)

\(=\left(x^2-2x\right)^2-17\left(x^2-2x\right)+72\)

\(=\left(x^2-2x-8\right)\left(x^2-2x-9\right)\)

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

x²-2xy+y²+3x-3y-10

= (x-y)²+3(x-y)-10

= [(x-y)²+5(x-y)]-[2(x-y)+10]

= (x-y)(x-y+5)-2(x-y+5)

= (x-y+5)(x-y-2)

Ta có: \(x^2-2xy+y^2+3x-3y-10\)

\(=\left(x-y\right)^2+3\left(x-y\right)-10\)

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

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

Phương trình thuần nhất đẳng cấp bậc 8 bạn nha :D

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

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

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

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

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

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

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

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

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

\(x^2-2x-24\)

\(=x^2-6x+4x-24\)

\(=x(x-6)+4(x-6)\)

\(=(x+4)(x-6)\)

\(x^2-2x-24\\ =x^2-2x+1-25\\ =\left(x-1\right)^2-5^2\\ =\left(x-1-5\right)\left(x-1+5\right)\\ =\left(x-6\right)\left(x+4\right)\)

\(x^2-2x-24=\left(x-6\right)\left(x+4\right)\)

\(5x^2-4\left(x^2-2x+1\right)-5=\left(5x^2-5\right)-4\left(x-1\right)^2=5\left(x^2-1\right)-4\left(x-1\right)^2=5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)^2=\left(x-1\right)\left[5\left(x+1\right)-4\left(x-1\right)\right]=\left(x-1\right)\left(5x+5-4x+4\right)=\left(x-1\right)\left(x+9\right)\)

\(= \)\(5x^2-4x^2+8x-4-5\)

\(=\)\(x^2+8x-9\)

\(=x^2+9x-x-9\)

\(=(x-1)(x+9)\)

\(5x^2-4\left(x^2-2x+1\right)-5\)

\(=5x^2-4x^2+8x-4-5\)

\(=x^2+8x-9\)

\(=\left(x+9\right)\left(x-1\right)\)