Giải phương trình sau 1909-x/91+1907-x/93+1905-x/95+1903-x/91+4=0

$\frac{1909-x}{91}+$ $\frac{1907-x}{93}+$ $\frac{1905-x}{95}+$ $\frac{1903-x}{97}+4=0$

$<=>$ $(\frac{1909-x}{91}+1)$ $(\frac{1907-x}{93}+1)+$ $(\frac{1905-x}{95}+1)+($ $\frac{1903-x}{97}+1)=0$

$<=>$$\frac{2000-x}{91}+$ $\frac{2000-x}{93}+$ $\frac{2000-x}{95}+$ $\frac{2000-x}{97}$

$<=>(2000-x)($$\frac{1}{91}+$ $\frac{1}{93}+$ $\frac{1}{95}+$ $\frac{1}{97})=0$

$<=>2000-x=0.(Vì: $ $\frac{1}{91}+$ $\frac{1}{93}+$ $\frac{1}{95}+$ $\frac{1}{97})$$\neq0)$ 

$<=>x=2000$