Ta có :

\(\begin{cases}\left|x+\frac{1}{1.2}\right|\ge0\\\left|x+\frac{1}{2.3}\right|\ge0\\...\\\left|x+\frac{1}{99.100}\right|\ge0\end{cases}\)\(\left(\forall x\right)\)

\(\Rightarrow100x>0\)

=> x > 0

=> \(\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+....+\left|x+\frac{1}{99.100}\right|\)

\(=x+\frac{1}{1.2}+x+\frac{1}{2.3}+.....+x+\frac{1}{99.100}=100x\)

\(\Rightarrow100x+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=100x\)

\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=0\)

Dễ thấy VT \(\ne\)VP

=> \(x\in\varnothing\)

Ta có: \(\left|x+\frac{1}{1.2}\right|\ge0;\left|x+\frac{1}{2.3}\right|\ge0;...;\left|x+\frac{1}{99.100}\right|\ge0\)

=> \(\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+...+\left|x+\frac{1}{99.100}\right|\ge0\)

=> \(100x\ge0\Rightarrow x\ge0\)

=> \(\left|x+\frac{1}{1.2}\right|=\left(x+\frac{1}{1.2}\right);\left|x+\frac{1}{2.3}\right|=\left(x+\frac{1}{2.3}\right);...;\left|x+\frac{1}{99.100}\right|=\left(x+\frac{1}{99.100}\right)\)=> \(\left(x+\frac{1}{1.2}\right)+\left(x+\frac{1}{2.3}\right)+...+\left(x+\frac{1}{99.100}\right)=100x\)

=> 99x + \(\frac{99}{100}\) = 100x

=> x = \(\frac{99}{100}\)

Này ko được chơi đểu nhá!!!

đề chưa đầy đủ

à đề thiếu tổng các giá trị tuyệt đối ở trên =100x

What wrong with you ?

Vì GTTĐ luôn lớn hơn hoặc bằng 0 với mọi x

\(\Rightarrow\left|x+\frac{1}{1\cdot2}\right|+\left|x+\frac{1}{2\cdot3}\right|+...+\left|x+\frac{1}{99\cdot100}\right|\ge0\)

\(\Rightarrow100x\ge0\)

\(\Rightarrow x\ge0\)

Từ điều kiện trên ta có :

\(x+\frac{1}{1\cdot2}+x+\frac{1}{2\cdot3}+...+x+\frac{1}{99\cdot100}=100x\)

\(50x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=100x\)

\(50x=1-\frac{1}{100}\)

\(50x=\frac{99}{100}\)

\(x=\frac{99}{5000}\)

Do \(\left|a\right|\ge0\forall a\) nên:

\(A=\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+...+\left|x+\frac{1}{99.100}\right|\ge0\forall x\)

\(\Leftrightarrow100x\ge0\) hay \(x\ge0\)

Do vậy ta có: \(A=\left(x+x+...+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)=100x\) ( 50 chữ số x)

\(\Leftrightarrow A=50x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=100x\)

\(\Leftrightarrow50x+\left(1-\frac{1}{100}\right)=100x\Leftrightarrow50x+\frac{99}{100}=100x\)

\(\Leftrightarrow50x=\frac{99}{100}\Leftrightarrow x=\frac{99}{100.50}=\frac{99}{5000}\)

tưởng là có 99x lận mà

a, \(\left(\frac{1}{2}\right)^x+\left(\frac{1}{2}\right)^{x+4}=17\)

\(\Rightarrow\frac{1}{2^x}+\frac{1}{2^x}\cdot\frac{1}{16}=17\)

\(\Rightarrow\frac{1}{2^x}\left(1+\frac{1}{16}\right)=17\)

\(\Rightarrow\frac{1}{2^x}\cdot\frac{17}{16}=17\)

\(\Rightarrow\frac{1}{2^x}=17:\frac{17}{16}=\frac{1}{16}=\frac{1}{2^4}\)

=> x = 4

b, Ta có: \(\left|x+\frac{1}{1.2}\right|\ge0;\left|x+\frac{1}{2.3}\right|\ge0;....;\left|x+\frac{1}{99.100}\right|\ge0\)

\(\Rightarrow\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+...+\left|x+\frac{1}{99.100}\right|\ge0\)

\(\Rightarrow100x\ge0\Rightarrow x\ge0\)

\(\Rightarrow x+\frac{1}{1.2}+x+\frac{1}{2.3}+...+x+\frac{1}{99.100}=100x\)

\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)=100x\)

\(\Rightarrow99x+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=100x\)

\(\Rightarrow100x-99x=1-\frac{1}{100}\)

\(\Rightarrow x=\frac{99}{100}\)

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x(x+1)}=\frac{2019}{2020}\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{2020}\)

\(\Rightarrow1-\frac{1}{x+1}=\frac{2019}{2020}\)

\(\Rightarrow\frac{1}{x+1}=1-\frac{2019}{2020}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2020}\)

\(\Rightarrow x+1=2020\Leftrightarrow x=2019\)

Vậy x = 2019

các giá trị tuyệt đối trên có tổng lớn hơn hoặc bằng 0(>=0)

=>100x>=0

=>x>=0 =>x+1/(1.2) >0 ;x+1/(2.3)>0;x+1/(3.4);.....;x+1/(99.100)>0

=> ta có thể phá dấu giá trị tuyệt đối 

=>100x=x+x+...+x(có 99. x)+(1/(1.2)+1/(2.3)+..+1/(99.100))

=>100x=99x+99/100

=>x=99/100

Áp dụng công thức: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

Ta có:

VT=\(x-\left(\left(1-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{3}\right)-...\left(\frac{1}{98}-\frac{1}{99}\right)-\left(\frac{1}{99}-\frac{1}{100}\right)\right)\)

=\(x-\frac{1}{100}\)

Dễ dàng tìm được 

\(x-\frac{1}{100}=\frac{1}{100}\) 

\(x=\frac{1}{50}\)