Respuesta :

gmany

[tex]\dfrac{k+3}{4k-2}\cdot(12k^2+2k-4)=\dfrac{k+3}{4k-2}\cdot(12k^2-6k+8k-4}\\\\=\dfrac{k+3}{4k-2}\cdot[3k(4k-2)+2(4k-2)]=\dfrac{k+3}{4k-2}\cdot(4k-2)(3k+2)\\\\=(k+3)(3k+2)[/tex]

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