Giải:
\(x^4y^4+4\)
\(=\left(x^2y^2\right)^2+2^2\)
\(=\left(x^2y^2+2\right)\left(x^2y^2-2\right)\)
Vậy ...
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\(x^4y^4+4=\left(x^2y^2\right)^2+2^2-4x^2y^2+4x^2y^2\\ =\left[\left(x^2y^2\right)^2+2^2+4x^2y^2\right]-4x^2y^2\\ =\left(x^2y^2+2\right)^2-\left(2xy\right)^2\\ =\left(x^2y^2-2xy+2\right)\left(x^2y^2+2xy+2\right)\)