Sửa đề: \(x^2+5xy+2xy+10y^2\)

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

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

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

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

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

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

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

\(x^4-5x^2y^2+4y^4\)

\(=\left(x^2\right)^2-2x^22y^2+\left(2y^2\right)^2-x^2y^2\)

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

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

Đề đúng không thế. Nếu đúng thì bài này phức tạp lắm

\(x^8+3x^3+1\)

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

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

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

\(=\left(x^4+1\right)\cdot\left(x^4+3\right)\)

sai rồi bạn

\(=x^2+2x\cdot\frac{1}{2}+\frac{1}{4}-\left(\frac{\sqrt{23}}{2}i\right)^2\)

\(=\left(x+\frac{1}{2}\right)^2\)\(-\left(\frac{\sqrt{23}}{2}i\right)^2\)

\(\left(x+\frac{1}{2}-\frac{\sqrt{23}}{2}i\right)\left(x+\frac{1}{2}+\frac{\sqrt[]{23}}{2}i\right)\)

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

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

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

\(x-\sqrt{x}-2\\ =x+\sqrt{x}-2\sqrt{x}-2\\ =\sqrt{x}\left(\sqrt{x}+1\right)-2\left(\sqrt{x}+1\right)\\ =\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)\)