We have: \(\dfrac{\left(4\times7+2\right)\left(6\times9+2\right)...\left(100\times103+2\right)}{\left(5\times8+2\right)\left(7\times10+2\right)...\left(99\times102+2\right)}=\dfrac{5\times6\times7\times8\times...\times101\times102}{6\times7\times8\times9\times...\times100\times101}=5\times102=510\)

Thank you !!!!!!!!! :D