Tìm x: $\frac{x-1}{8}$ + $\frac{x-1}{9}$ = $\frac{x-1}{10}$ +$\frac{x-1}{11}$

Em tham khảo:

  $\frac{x-1}{8}+$$\frac{x-1}{9}=$$\frac{x-1}{10}+$$\frac{x-1}{11}$    

⇔ $\frac{x-1}{8}+$$\frac{x-1}{9}-$$\frac{x-1}{10}-$$\frac{x-1}{11}=0$    

⇔$(x-1)$.($\frac{1}{8}+$$\frac{1}{9}-$$\frac{1}{10}-$$\frac{1}{11}$) $=0$    

Vì $\frac{1}{8}+$$\frac{1}{9}-$$\frac{1}{10}-$$\frac{1}{11}>0$

⇔$x-1=0$

⇔$x=1$

HỌC TỐT 

@Chin………..