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

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

\(\Rightarrow11x\ge0\)

\(\Rightarrow x\ge0\)

Với \(x\ge0\) ta có:

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

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

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

10 số x

\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=11x\)

\(\Rightarrow1-\frac{1}{11}=11x-10x\)

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

Vậy \(x=\frac{10}{11}\)

Mày nhìn cái chóa j

Ta có:

\(\left\{{}\begin{matrix}\left|x+\frac{1}{2}\right|\ge0\\\left|x+\frac{1}{6}\right|\ge0\\...\\\left|x+\frac{1}{110}\right|\ge0\end{matrix}\right.\)

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

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

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

=\(x+\frac{1}{2}+x+\frac{1}{6}+...+x+\frac{1}{110}\)

\(=10x+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)\)

Đặt \(A=\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\)

\(\Rightarrow A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{11-10}{10.11}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)

\(\Rightarrow A=1-\frac{1}{11}=\frac{10}{11}\)

\(\Rightarrow10x+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)=10x+A=10x+\frac{10}{11}=11x\)

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

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

Ta có:

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

\(\Rightarrow VT\ge0\Rightarrow VP\ge0\)

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

\(\Rightarrow\left(x+...+x\right)+\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{110}\right)=11x\)

\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10\cdot11}\right)=11x\)

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

\(\Rightarrow10x+\left(1-\frac{1}{11}\right)=11x\)

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

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

mk viết vội nên nhầm dòng thứ 4 từ trên xuống, bn sửa 1/3 thành 1/6 nhé

kq vẫn đúng đấy

vì sao biết đc là 10x 

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.....

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

Với mọi x ta có:

+) \(\left\{{}\begin{matrix}\left|x+\frac{1}{2}\right|\ge0\\\left|x+\frac{1}{6}\right|\ge0\\.........\\\left|x+\frac{1}{110}\right|\ge0\end{matrix}\right.\)

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

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

\(\Rightarrow11x\ge0\)

\(\Rightarrow x\ge0.\)

Với \(x\ge0\) thì:

\(\left\{{}\begin{matrix}\left|x+\frac{1}{2}\right|=x+\frac{1}{2}\\\left|x+\frac{1}{6}\right|=x+\frac{1}{6}\\..........\\\left|x+\frac{1}{110}\right|=x+\frac{1}{110}\end{matrix}\right.\)

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

\(\Rightarrow11x+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)=11x\)

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

\(\Rightarrow\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}=0x\) (vô lí).

\(\Rightarrow x\in\varnothing.\)

Vậy không tồn tại giá trị của x thỏa mãn yêu cầu đề bài.

Chúc bạn học tốt!

mình sửa đề chút nhé + \(\left|x+\frac{1}{110}\right|=11x\)

Vế trái lớn hơn hoặc bằng 0 nên 11x lớn hơn hoặc bằng 0.

\(\Rightarrow x\ge0\)

Do vậy chỉ cần bỏ dấu giá trị tuyệt đối là tính được.

Kết quả cuối cùng được \(x=\frac{10}{11}\)

\(\Leftrightarrow8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)\left[\left(x^2+\frac{1}{x^2}\right)-\left(x+\frac{1}{x}\right)^2\right]=\left(x+4\right)^2.ĐKXĐ:x\ne0\)

\(\Leftrightarrow8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)\left(x^2+\frac{1}{x^2}-x^2-2-\frac{1}{x^2}\right)=\left(x+4\right)^2\)

\(\Leftrightarrow8\left(x+\frac{1}{x}\right)^2-8\left(x^2+\frac{1}{x^2}\right)=\left(x+4\right)^2\)

\(\Leftrightarrow8\left[\left(x+\frac{1}{x}\right)^2-\left(x^2+\frac{1}{x^2}\right)\right]=\left(x+4\right)^2\)

\(\Leftrightarrow8\left(x^2+2+\frac{1}{x^2}-x^2+\frac{1}{x^2}\right)=\left(x+4\right)^2\)

\(\Leftrightarrow16=\left(x+4\right)^2\)

\(\Leftrightarrow x^2+8x+16=16\)

\(\Leftrightarrow x^2+8x=0\)

\(\Leftrightarrow x\left(x+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\left(l\right)\\x=-8\left(n\right)\end{cases}}\)

V...\(S=\left\{-8\right\}\)

^^

bạn ghi sai đề ở chỗ \(\left(x+\frac{1}{x}\right)^2\)chứ ko phải \(\left(x+\frac{1}{x^2}\right)^2\)nhé