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\(x^2-y^2=\left(x-y\right)\left(x+y\right)\)\(=>\)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!!
\(=x^2-6x+9-2=\left(x-3\right)^2-2=\left(x-3-\sqrt{2}\right)\left(x-3+\sqrt{2}\right)\)
\(=x^2+8x-x-8=x\left(x+8\right)-\left(x+8\right)=\left(x+8\right)\left(x-1\right)\)
Tham khảo:https://hoc247.net/hoi-dap/toan-8/phan-tich-da-thuc-x-7-x-2-1-thanh-nhan-tu-faq417522.html
\(=x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2+x^2-x^2+x-x+1\\ =\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+\left(x^4+x^3+x^2\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\)
\(x^3-x^2y+3x-3y\)
\(=x^2\left(x-y\right)+3\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2+3\right)\)
\(=x^2\left(x-y\right)+3\left(x-y\right)=\left(x^2+3\right)\left(x-y\right)\)
\(x^2-y^2+x-y\)
\(=\left(x^2-y^2\right)+\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y\right)+\left(x+y\right)\)
\(=\left(x-y+1\right)\left(x+y\right)\)