\(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{6}\right|+...+\left|x+\dfrac{1}{110}\right|=11x\left(đk:x\ge0\right)\)

\(\Leftrightarrow x+\dfrac{1}{2}+x+\dfrac{1}{6}+x+\dfrac{1}{12}+...+x+\dfrac{1}{110}=11x\)

\(\Leftrightarrow10x+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{10.11}\right)=11x\)

\(\Leftrightarrow x=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{10}-\dfrac{1}{11}\)

\(\Leftrightarrow x=1-\dfrac{1}{11}=\dfrac{10}{11}\left(tm\right)\)

Ta có: \(\left|x+\frac{1}{2}\right|\ge0\left|x+\frac{1}{6}\right|\ge0;...;\left|x+\frac{1}{110}\ge0\right|\)

\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{100}\right|\ge0\)

\(\Rightarrow11x\ge0\Rightarrow x\ge0\)

\(\Rightarrow x+\frac{1}{2}>0;x+\frac{1}{6}>0;...;x+\frac{1}{100}>0\)

\(\Rightarrow\left|x+\frac{1}{2}\right|=x+\frac{1}{2};\left|x+\frac{1}{6}\right|=x+\frac{1}{6};...;\left|x+\frac{1}{100}\right|=x+\frac{1}{110}\)

\(\Rightarrow\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+...+\left(x+\frac{1}{110}\right)=11x\)

\(\Rightarrow10x+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)=11x\)

\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=11x\)

\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)=11x\)

\(\Rightarrow10x+\frac{10}{11}=11x\)

\(\Rightarrow x=\frac{10}{11}\)

vì |x+1/2| ; |x+1/6| ; ............ ; |x+110| lớn hơn hoặc bằng 0=> 11x lớn hớn hoặc bằng 0=> x lớn hớn hoặc bằng 0

=>x+1/2 ; x+1/6 ; ............ ; x+110 lớn hơn hoặc bằng 0

ta có: x+1/2+x+1/6+x+1/12+...+x+1/110=11x

(x+x+...+x)+(1/1.2+1/2.3+1/3.4+...+1/10.11)=11x

10x+(1-1/10)=11x

x= 1/9

à mình bỏ dấu" | " vì khi mà lớn hơn hoặc bằng 1 rồi thfi bỏ ra nó vẫn có giá trị bằng giá trị trị lúc ban đầu

\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+\left|x+\frac{1}{12}\right|+...\left|x+\frac{1}{110}\right|=11x\)

\(\Leftrightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+\left|x+\frac{1}{12}\right|+...\left|x+\frac{1}{110}\right|\ge0\)

\(\rightarrow11x\ge0\rightarrow x\ge0\)

\(\text{Ta có:}\)

\(x+\frac{1}{2}+...+x+\frac{1}{110}=11x\)

\(\rightarrow10x+\frac{10}{11}=11x\)

\(\rightarrow x=\frac{10}{11}\)

Với \(\forall x\) ta có :

+) \(\left|x+\dfrac{1}{2}\right|\ge0\)

+) \(\left|x+\dfrac{1}{6}\right|\ge0\)

..........................

+) \(\left|x+\dfrac{1}{110}\right|\ge0\)

\(\Leftrightarrow\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{6}\right|+.........+\left|x+\dfrac{1}{110}\right|\ge0\)

Mà \(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{6}\right|+........+\left|x+\dfrac{1}{110}\right|=11x\)

\(\Leftrightarrow11x\ge0\)

\(\Leftrightarrow x\ge0\)

Với \(x\ge0\) thì :

+) \(\left|x+\dfrac{1}{2}\right|=x+\dfrac{1}{2}\)

+) \(\left|x+\dfrac{1}{6}\right|=x+\dfrac{1}{6}\)

.....................................

+) \(\left|x+\dfrac{1}{110}\right|=x+\dfrac{1}{110}\)

\(\Leftrightarrow x+\dfrac{1}{2}+x+\dfrac{1}{6}+......+x+\dfrac{1}{110}=11x\)

\(\Leftrightarrow11x+\left(\dfrac{1}{2}+\dfrac{1}{6}+........+\dfrac{1}{110}\right)=11x\)

\(\Leftrightarrow0x=\dfrac{1}{2}+\dfrac{1}{6}+....+\dfrac{1}{110}\) (vô lí)

\(\Leftrightarrow x\in\varnothing\)

Ta có : \(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+\left|x+\frac{1}{12}\right|+...+\left|x+\frac{1}{110}\right|\ge0\forall x\)

=> 11x \(\ge\)0

=> x  \(\ge\)0 

Khi đó \(\orbr{\begin{cases}x+\frac{1}{2}+x+\frac{1}{6}+x+\frac{1}{12}+...+x+\frac{1}{110}=11x\left(10\text{ số hạng x }\right)\\x+\frac{1}{2}+x+\frac{1}{6}+x+\frac{1}{12}+...+x+\frac{1}{110}=-11x\left(10\text{ số hạng x}\right)\end{cases}}\)

=> \(\orbr{\begin{cases}10x+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)=11x\\10x+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)=-11x\end{cases}}\)

=> \(\orbr{\begin{cases}10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)=11x\\10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)=-11x\end{cases}}\)

=> \(\orbr{\begin{cases}10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=11x\\10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=-11x\end{cases}}\)

=> \(\orbr{\begin{cases}10x+\left(1-\frac{1}{11}\right)=11x\\10x+\left(1-\frac{1}{11}\right)=-11x\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{10}{11}\\21x=-\frac{10}{11}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{10}{11}\left(\text{tm}\right)\\x=-\frac{10}{231}\left(\text{loại}\right)\end{cases}}}\)

Vậy \(x=\frac{10}{11}\)

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