Ta có: \(\left|x+\frac{1}{2}\right|\ge0;\left|x+\frac{1}{6}\right|\ge0;\left|x+\frac{1}{12}\right|\ge0;...;\left|x+\frac{1}{110}\right|\ge0\)
=> VT \(\ge\)0
=>VP \(\ge\)0 => 11x \(\ge\)0 => x \(\ge\)0.
=> \(\left|x+\frac{1}{2}\right|=x+\frac{1}{2};\left|x+\frac{1}{6}\right|=x+\frac{1}{6};\left|x+\frac{1}{12}\right|=x+\frac{1}{12};...;\left|x+\frac{1}{110}\right|=x+\frac{1}{110}\)
Phương trình <=> \(x+\frac{1}{2}+x+\frac{1}{6}+x+\frac{1}{12}+...+x+\frac{1}{110}=11x\)
<=> \(\left(x+x+x+...+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)=11x\)
<=> \(10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=11x\)
<=> \(1-\frac{1}{11}=11x-10x\)
<=> \(\frac{10}{11}=x\)
<=> \(x=\frac{10}{11}\left(tm\right)\)
Bởi vì
\(\frac{1}{2}=\frac{1}{1.2};\frac{1}{6}=\frac{1}{2.3};...;\frac{1}{110}=\frac{1}{10.11}\)
nên từ \(\frac{1}{2}\)đến \(\frac{1}{110}\)chỉ có 10 số
nên chỉ có 10 x
Ta có : \(|x+\frac{1}{2}|\ge0;|x+\frac{1}{6}|\ge0;....;|x+\frac{1}{110}|\ge0\)
\(\Rightarrow|x+\frac{1}{2}|+|x+\frac{1}{6}|+...+|x+\frac{1}{110}|\ge0\)
\(\Rightarrow11x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow x+\frac{1}{2}>0;x+\frac{1}{6}>0;....;x+\frac{1}{110}>0\)
\(\Rightarrow|x+\frac{1}{2}|=x+\frac{1}{2};|x+\frac{1}{6}|=x+\frac{1}{6};....;|x+\frac{1}{110}|=x+\frac{1}{110}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+...+\left(x+\frac{1}{110}\right)=11x\)
\(\Rightarrow10x+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)=11x\)
\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=11x\)
\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{10}-\frac{1}{11}\right)=11x\)
\(\Rightarrow10x+\frac{1}{10}=11x\)
\(\Rightarrow x=\frac{10}{11}\)
Vậy.....