Ta có : \(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}=\frac{x+1}{5}+\frac{x+1}{6}\)
\(\Rightarrow\)\(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0\)
\(\Rightarrow\)\(\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)
Mà : \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)\ne0\)
Nên x + 1 = 0
=> x = -1
Vậy x = -1
\(\left(\frac{x+1}{2}\right)+\left(\frac{x+1}{3}\right)+\left(\frac{x+1}{4}\right)=\left(\frac{x+1}{5}\right)+\left(\frac{x+1}{6}\right)\)
\(\Leftrightarrow\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0\)
\(\Leftrightarrow x+1\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)
\(\Leftrightarrow x+1=0\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
