Question

Giải phương trình sau:
\(\frac{x-2004}{15}+\frac{x-1995}{12}+\frac{x-1989}{10}+\frac{x-1987}{8}=10.\)

Answer 1

Giải phương trình sau:
\(\frac{x-2004}{15}\)+\(\frac{x-1995}{12}\)+\(\frac{x-1989}{10}\)+\(\frac{x-1987}{8}\)=\(10\)

⇔\(\frac{\left(x-2004\right).40}{600}\) +\(\frac{\left(x-1995\right).50}{600}\)+\(\frac{\left(x-1989\right).60}{600}\)+\(\frac{\left(x-1987\right).75}{600}\)=\(\frac{10.600}{600}\)

⇔\(\frac{40x-80160}{600}\) + \(\frac{50x-99750}{600}\) +\(\frac{60x-119340}{600}\) +\(\frac{75x-149025}{600}\)=\(\frac{6000}{600}\)

➞ \(40x-80160+50x-99750+60x-119340+75x-149025=6000\)⇔\(225x=\)\(6000+80160+99750+119340+149025\)

⇔\(225x=454275\)

⇔\(x=2019\)

Answer 2

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Answer 3

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