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

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

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

=> x = -1 

sửa lại đề : tìm x biết : \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

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

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

dễ thấy \(\frac{1}{10}>\frac{1}{11}>\frac{1}{12}>\frac{1}{13}>\frac{1}{14}\)nên biểu thức trong dấu ngoặc thứ hai khác 0

\(\Rightarrow x+1=0\)

\(\Rightarrow x=-1\)

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

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

dễ thấy \(\frac{1}{10}>\frac{1}{11}>\frac{1}{12}>\frac{1}{13}>\frac{1}{14}\)nên biểu thức trong dấu ngoặc thứ hai khác 0

\(\Rightarrow x+1=0\)

\(\Rightarrow x=-1\)

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

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

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

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

=> x= -1

Vậy x= -1

\(a)\)

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

\(\Leftrightarrow x=\frac{11}{12}-\frac{1}{10}-\frac{11}{11}\)

\(\Leftrightarrow x=\frac{-11}{60}\)

\(b)\)

\(-\left|x-\frac{1}{2}\right|=\frac{1}{2}\)

\(\Leftrightarrow\left|x-\frac{1}{2}\right|=\frac{-1}{2}\) (Vô lý)

Vậy \(x\in\varnothing\)