\(\frac{x^2+4y^2-4xy-4}{2x^2-4xy+4x}=\frac{\left(x^2-4xy+4y^2\right)-4}{2x.\left(x-2y+2\right)}.\)
\(=\frac{\left(x-2y\right)^2-4}{2x.\left(x-2y+2\right)}=\frac{\left(x-2y+2\right).\left(x-2y-2\right)}{2x.\left(x-2y+2\right)}\)
\(=\frac{x-2y-2}{2x}\)
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