a) $\frac{x}{2012}$+ $\frac{x+1}{2013}$ +$\frac{x+2}{2014}$+ $\frac{x+3}{2015}$+ $\frac{x+4}{2016}$ =5
b)x
$\frac{x-90}{10}$ +$\frac{x-76}{12}$+ $\frac{x-58}{14}$+ $\frac{x-36}{16}$+ $\frac{x-15}{17}$=15
c) $\frac{x^2-10x-29}{1971}$ + $\frac{x^2-10x-27}{1973}$ =$\frac{x^2-10x-1971}{29}$ +$\frac{x^2-10x-1073}{27}$
Giải các PT trên
a, ⇔$(\frac{x}{2012}-1)+($ $\frac{x+1}{2013}-1)+($ $\frac{x+2}{2014}-1)+($ $\frac{x+3}{2015}-1)+($ $\frac{x+4}{2016}-1)=5-5$ =>$\frac{x-2012}{2012}+$ $\frac{x-2012}{2013}+$ $\frac{x-2012}{2014}+$ $\frac{x-2012}{2015}+$ $\frac{x-2012}{2016}=0$ =>$(x-2012)($$\frac{1}{2012}+$ $\frac{1}{2013}+$ $\frac{1}{2014}+$ $\frac{1}{2015}+$ $\frac{1}{2016})=0$
⇔$x-2012=0.VÌ: $$\frac{1}{2012}+$ $\frac{1}{2013}+$ $\frac{1}{2014}+$ $\frac{1}{2015}+$ $\frac{1}{2016}$$\neq0$
⇔$x=2012$
b, ⇔$(\frac{x-90}{10}-1)+$ $(\frac{x-76}{12}-2)+($ $\frac{x-58}{14}-3)+($ $\frac{x-36}{16}-4)+($ $\frac{x-15}{17}-5)=15-15$ =>$\frac{x-100}{10}+$ $\frac{x-100}{12}+$ $\frac{x-100}{14}+$ $\frac{x-100}{16}+$ $\frac{x-100}{17}=0$ =>$(x-100)(\frac{1}{10}+$ $\frac{1}{12}+$ $\frac{1}{14}+$ $\frac{1}{16}+$ $\frac{1}{17})=0$
⇔$x-100=0.Vì: $$\frac{1}{10}+$ $\frac{1}{12}+$ $\frac{1}{14}+$ $\frac{1}{16}+$ $\frac{1}{17}$$\neq0$
c, ⇔$(\frac{x^2-10x-29}{1971}-1)+($ $\frac{x^2-10x-27}{1973}-1)-$ $(\frac{x^2-10x-1971}{29}-1)-$ $(\frac{x^2-10x-1973}{27}-1)=0$ =>$\frac{x^2-10x-200}{1971}+$ $\frac{x^2-10x-2000}{1973}-$ $\frac{x^2-10x-2000}{29}-$ $\frac{x^2-10x-2000}{27}=0$ =>$(x^2-10x-2000)(\frac{1}{1971}+$ $\frac{1}{1973}-$ $\frac{1}{29}-$ $\frac{1}{27})=0$
⇔$x^2-10x-2000=0.Vì: $$\frac{1}{1971}+$ $\frac{1}{1973}-$ $\frac{1}{29}-$ $\frac{1}{27}$$\neq0$
⇔$x^2 – 50x + 40x – 2000 = 0$
⇔$(x-50)(x+40)=0$
⇔$x=50; x=-40$
Vậy S={50;-40}