\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\Rightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Ta có:\(\hept{\begin{cases}\frac{1}{10}>\frac{1}{14}\\\frac{1}{11}>\frac{1}{15}\end{cases}\Rightarrow\frac{1}{10}+\frac{1}{11}-\frac{1}{14}-\frac{1}{15}+\frac{1}{12}>0}\)

\(\Rightarrow x+1=0\Rightarrow x=-1\)

Cho mình hỏi mình làm thế này có đúng ko ? Nếu đúng k mình nha !!

(x +1) / 10 + (x + 1) / 11 + (x + 1) / 12 = (x + 1) / 13 + (x + 1) / 14 
=>  (x + 1)( 1/10 + 1/11 + 1/12 - 1/13 - 1/14) = 0 
=> x + 1 = 0   hoặc 1/10 + 1/11 + 1/12 - 1/13 - 1/14 = 0 -- vô lí do VT # 0 
=>  x = -1 
vậy x = - 1

\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

=> \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

=> \(\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\)

=> x+1=0

=> x=-1

\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

=> \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

= (x + 1) (1/10 + 1/11 + 1/12 - 1/13 - 1/14 ) = 0

vi  (1/10 + 1/11 + 1/12 - 1/13 - 1/14 ) \(\ne\)0

=> x + 1 = 0 

=> x = - 1

= (x + 1) (1/10 + 1/11 + 1/12 + 1/13 + 1/14 ) = 0

=> x + 1 = 0 

=> x = - 1