Bài 5:
a, $\frac{59-x}{41}$+ $\frac{57-x}{43}$+ $\frac{55-x}{45}$+ $\frac{53-x}{47}$ +$\frac{51-x}{49}$= $-5$
⇔ $\frac{59-x}{41}+1+ \frac{57-x}{43}+1+ \frac{55-x}{45}+1 +\frac{53-x}{47}+1 +\frac{51-x}{49}+1= 0$
⇔ $\frac{100-x}{41}$+ $\frac{100-x}{43}$+ $\frac{100-x}{45}$+ $\frac{100-x}{47}$ +$\frac{100-x}{49}$
⇔ $( 100-x).( \frac{1}{41}+\frac{1}{43}+\frac{1}{45}+\frac{1}{47}+\frac{1}{49})=0$
⇔ $100-x= 0$
⇔ $x= 100$
b, $\frac{x-5}{1990}+\frac{x-15}{1980}+\frac{x-25}{1970}= \frac{x-1990}{5}+\frac{x-1980}{15}+\frac{x-1970}{25}$
⇔ $\frac{x-5}{1990}+1+\frac{x-15}{1980}+1+\frac{x-25}{1970}+1= \frac{x-1990}{5}+1+\frac{x-1980}{15}+1+\frac{x-1970}{25}+1$
⇔ $\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}= \frac{x-1995}{5}+\frac{x-1995}{15}+\frac{x-1995}{25}$
⇔ $( x-1995).( \frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}-\frac{1}{5}-\frac{1}{15}-\frac{1}{25})=0$
⇔ $x-1995= 0$
⇔ $x= 1995$
c, $\frac{x+14}{86} +\frac{x+15}{85} +\frac{x+16}{84} +\frac{x+17}{83} +\frac{x+116}{4} =0$
⇔ $\frac{x+14}{86}+1 +\frac{x+15}{85}+1 +\frac{x+16}{84}+1 +\frac{x+17}{83}+1 +\frac{x+116}{4}-4 =0$
⇔ $\frac{x+100}{86} +\frac{x+100}{85} +\frac{x+100}{84} +\frac{x+100}{83} +\frac{x+100}{4} =0$
⇔ $( x+100).( \frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}= 0$
⇔ $x+100= 0$
⇔ $x= -100$
