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

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:

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

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

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

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

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

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

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

=> 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 + \(\left|x+\frac{1}{110}\right|=11x\)

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

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 \(x=\frac{10}{11}\)

Bài 1:

\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{101}\right|=101x\)

Ta thấy:

\(VT\ge0\Rightarrow VP\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+...+\left(x+\frac{1}{101}\right)=101x\)

\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{101}\right)=0\)

\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=0\)

\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)=0\)

\(\Rightarrow10x+\left(1-\frac{1}{11}\right)=0\)

\(\Rightarrow10x+\frac{10}{11}=0\)

\(\Rightarrow10x=-\frac{10}{11}\Rightarrow x=-\frac{1}{11}\)(loại,vì x\(\ge\)0)

Bài 2:

Ta thấy: \(\begin{cases}\left(2x+1\right)^{2008}\ge0\\\left(y-\frac{2}{5}\right)^{2008}\ge0\\\left|x+y+z\right|\ge0\end{cases}\)

\(\Rightarrow\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|\ge0\)

Mà \(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)

\(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)

\(\Rightarrow\begin{cases}\left(2x+1\right)^{2008}=0\\\left(y-\frac{2}{5}\right)^{2008}=0\\\left|x+y+z\right|=0\end{cases}\)\(\Rightarrow\begin{cases}2x+1=0\\y-\frac{2}{5}=0\\x+y+z=0\end{cases}\)

\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\x+y+z=0\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{2}+\frac{2}{5}+z=0\end{cases}\)

\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{10}=-z\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{1}{10}\end{cases}\)

Bài 3:

a)\(2009-\left|x-2009\right|=x\)

\(\Rightarrow\left|x-2009\right|=2009-x\)

\(\Rightarrow\left|x-2009\right|=-\left(x-2009\right)\)

Vì GTTĐ của số âm bằng số đối của nó

\(\Rightarrow x-2009\le0\)

\(\Rightarrow x\le2009\)

Vậy với mọi \(x\le2009\) đều thỏa mãn

b)\(\left|3x+2\right|=\left|5x-3\right|\)

\(\Rightarrow3x+2=5x-3\) hoặc \(3x+2=3-5x\)

\(\Rightarrow2x=5\) hoặc \(8x=1\)

\(\Rightarrow x=\frac{5}{2}\) hoặc \(x=\frac{1}{8}\)

Ta có vế trái của pt luôn \(\ge0\)

Do đó : \(11x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\hept{\begin{cases}\left|x+\frac{1}{2}\right|=x+\frac{1}{2}\\...\\\left|x+\frac{1}{110}\right|=x+\frac{1}{110}\end{cases}}\)

Khi đó pt trở thành :

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

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

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

\(\Leftrightarrow x=1-\frac{1}{11}=\frac{10}{11}\) ( thỏa mãn )

Vậy : pt đã cho có nghiệm \(S=\left\{\frac{10}{11}\right\}\)

Dễ thấy \(VT>0\forall x\)

\(\Rightarrow11x>0\Rightarrow x>0\)

Phương trình trở thành \(10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=11x\)

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

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

Vậy \(x=\frac{10}{11}\)