c: \(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{6}\right|+...+\left|x+\dfrac{1}{110}\right|=11x\)
mà \(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{6}\right|+...+\left|x+\dfrac{1}{110}\right|>=0\)
nên 11x>=0
=>x>=0
Phương trình sẽ trở thành:
\(x+\dfrac{1}{2}+x+\dfrac{1}{6}+...+x+\dfrac{1}{110}=11x\)
=>\(x+\dfrac{1}{1\cdot2}+x+\dfrac{1}{2\cdot3}+...+x+\dfrac{1}{10\cdot11}=11x\)
=>\(11x=10x+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{10}-\dfrac{1}{11}\)
=>\(11x=10x+1-\dfrac{1}{11}=10x+\dfrac{10}{11}\)
=>\(x=\dfrac{10}{11}\left(nhận\right)\)