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