Đa thức này ko phân tích thành nhân tử được

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

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

\(x^4\ge0;x^2\ge0;4>0\Rightarrow x^4+x^2+4>0\)

đề lỗi rồi

ghi lộn chỉ có x^4+x^2+4 thôi

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

\(\dfrac{1}{4}x^2+2xy+4y^2=\left(\dfrac{1}{2}x+2y\right)^2\)

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

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

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

       \(x^4+2002x^2-2001x+2002\)

\(=x^4+2002x^2+x-2002x+2002\)

\(=\left(x^4+x\right)+\left(2002x^2-2002x+2002\right)\)

\(=x\left(x^3+1\right)+2002\left(x^2-x+1\right)\)

\(=x\left(x+1\right)\left(x^2-x+1\right)+2002\left(x^2-x+1\right)\)

\(=\left(x^2-x+1\right)\left[x\left(x+1\right)+2002\right]\)

\(=\left(x^2-x+1\right)\left(x^2+x+2002\right)\)

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

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

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

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

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