\(x^2-y^2=\left(x-y\right)\left(x+y\right)\)\(=>\)Hằng đẳng thức Hiệu hai bình phương 

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

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

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

\(=\left(x-y\right).\left(x+y\right)+\left(x+y\right)\)

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

P/s: Tham khảo nha!!

\(=x^2-6x+9-2=\left(x-3\right)^2-2=\left(x-3-\sqrt{2}\right)\left(x-3+\sqrt{2}\right)\)

\(=\left(x^2-6x+9\right)-2=\left(x-3\right)^2-\sqrt{2^2}=\left(x-3-\sqrt{2}\right)\left(x-3+\sqrt{2}\right)\)

=x2−6x+9−2=(x−3)2−2=(x−3−√2)(x−3+√2)

\(=x^2+8x-x-8=x\left(x+8\right)-\left(x+8\right)=\left(x+8\right)\left(x-1\right)\)

\(x^2+8x-x-8=8\left(x-1\right)+x\left(x-1\right)=\left(8+x\right)\left(x-1\right)\)

   x²+7x-8
= x²-x+8x-8
= (x²-x)+(8x-8)
= x(x-1)+8(x-1)
= (x-1)(x+8)

Tham khảo:https://hoc247.net/hoi-dap/toan-8/phan-tich-da-thuc-x-7-x-2-1-thanh-nhan-tu-faq417522.html

\(=x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2+x^2-x^2+x-x+1\\ =\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+\left(x^4+x^3+x^2\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\)

Lời giải:
$x^2+x-6=(x^2-2x)+(3x-6)=x(x-2)+3(x-2)=(x-2)(x+3)$

\(4a^2b^2-\left(a^2+b^2-c^2\right)^2\)

\(=4a^2b^2-2ab\left(a^2+b^2-c^2\right)+2ab\left(a^2+b^2-c^2\right)-\left(a^2+b^2-c^2\right)^2\)

\(=2ab\left[2ab-\left(a^2+b^2-c^2\right)\right]+\left(a^2+b^2-c^2\right)\left[2ab-\left(a^2+b^2-c^2\right)\right]\)

\(=\left(2ab+a^2+b^2-c^2\right)\left(2ab-a^2-b^2+c^2\right)\)

\(=\left(a^2+ab+ab+b^2-c^2\right)\left[c^2-\left(a^2-ab-ab+b^2\right)\right]\)

\(=\left[a\left(a+b\right)+b\left(a+b\right)-c^2\right]\left[c^2-\left(a\left(a-b\right)-b\left(a-b\right)\right)\right]\)

\(=\left[\left(a+b\right)^2-c^2\right]\left[c^2-\left(a-b\right)^2\right]\)

\(=\left[\left(a+b\right)^2-c\left(a+b\right)+c\left(a+b\right)-c^2\right]\left[c^2+c\left(a-b\right)-c\left(a-b\right)-\left(a-b\right)^2\right]\)

\(=\left[\left(a+b\right)\left(a+b-c\right)+c\left(a+b-c\right)\right]\left[c\left(c+a-b\right)-\left(a-b\right)\left(c+a-b\right)\right]\)

\(=\left(a+b+c\right)\left(a+b-c\right)\left(c+a-b\right)\left(c-a+b\right)\)

bạn có thể viết rõ hơn ko

x^4+x^2y^2+y^4

=x^4+2x^2y^2+y^4-x^2y^2

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

=(x^2-xy+y^2)(x^2+xy+y^2)

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

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

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

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

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