Ta có

\(\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+...+\left|x+\frac{1}{99.100}\right|=100x\)
\(\left|x+x+...x\right|+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)=100x\)
\(\left|99x\right|+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=100x\)
\(\left|99x\right|+\left(\frac{1}{1}-\frac{1}{100}\right)=100x\)
\(\left|99x\right|+\frac{99}{100}=100x\)
Sau đó tự biến đổi nha! Mik chỉ giải tới đó thôi vì mới lớp 6 à!

\(\left|x+\frac{1}{1\cdot2}\right|+\left|x+\frac{1}{2\cdot3}\right|+\left|x+\frac{1}{3\cdot4}\right|+...+\left|x+\frac{1}{99\cdot100}\right|=100x\)

có :

\(\left|x+\frac{1}{1\cdot2}\right|;\left|x+\frac{1}{2\cdot3}\right|;\left|x+\frac{1}{3\cdot4}\right|;...;\left|x+\frac{1}{99\cdot100}\right|\ge0\)

\(\Rightarrow\left|x+\frac{1}{1\cdot2}\right|+\left|x+\frac{1}{2\cdot3}\right|+\left|x+\frac{1}{3\cdot4}\right|+...+\left|x+\frac{1}{99\cdot100}\right|\ge0\)

\(\Rightarrow100x\ge0\)

\(\Rightarrow x\ge\frac{0}{100}\)

\(\Rightarrow x\ge0\)

\(\Rightarrow\left|x+\frac{1}{1\cdot2}\right|+\left|x+\frac{1}{2\cdot3}\right|+\left|x+\frac{1}{3\cdot4}\right|+...+\left|x+\frac{1}{99\cdot100}\right|\)

\(=x+\frac{1}{1\cdot2}+x+\frac{1}{2\cdot3}+x+\frac{1}{3\cdot4}+...+x+\frac{1}{99\cdot100}\)

bước này tự lm tp

a: \(A=\dfrac{1}{x-1}\cdot5\sqrt{3}\cdot\left|x-1\right|\cdot\sqrt{x-1}\)

\(=\dfrac{5\sqrt{3}}{x-1}\cdot\left(x-1\right)\cdot\sqrt{x-1}=5\sqrt{3}\cdot\sqrt{x-1}\)

b: \(B=10\sqrt{x}-3\cdot\dfrac{10\sqrt{x}}{3}-\dfrac{4}{x}\cdot\dfrac{x\sqrt{x}}{2}\)

\(=10\sqrt{x}-10\sqrt{x}-\dfrac{4\sqrt{x}}{2}=-2\sqrt{x}\)

c: \(C=x-4+\left|x-4\right|\)

=x-4+x-4

=2x-8

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

\(\dfrac{1}{5}\left(x+\dfrac{1}{5}\right)+\dfrac{2}{5}\left(x+\dfrac{5}{3}\right)=\dfrac{98}{75}\\ =>\dfrac{1}{5}x+\dfrac{1}{25}+\dfrac{2}{5}x+\dfrac{2}{3}=\dfrac{98}{75}\\ =>\dfrac{3}{5}x=\dfrac{98}{75}-\dfrac{2}{3}-\dfrac{1}{25}=\dfrac{3}{5}\\ =>x=1\)

\(\dfrac{1}{5}\left(x+\dfrac{1}{5}\right)+\dfrac{2}{5}\left(x+\dfrac{5}{3}\right)=\dfrac{98}{75}\\ \Rightarrow\dfrac{1}{5}x+\dfrac{1}{25}+\dfrac{2}{5}x+\dfrac{2}{3}=\dfrac{98}{75}\\ \Rightarrow\left(\dfrac{1}{5}x+\dfrac{2}{5}x\right)+\left(\dfrac{1}{25}+\dfrac{2}{3}\right)=\dfrac{98}{75}\\ \Rightarrow\dfrac{3}{5}x+\dfrac{53}{75}=\dfrac{98}{75}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{98}{75}-\dfrac{53}{75}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{45}{75}=\dfrac{3}{5}\\ \Rightarrow x=\dfrac{3}{5}:\dfrac{3}{5}\\ \Rightarrow x=1\)

\(\dfrac{1}{5}\)x(X+\(\dfrac{1}{5}\))+\(\dfrac{2}{5}\)x(X+\(\dfrac{5}{3}\))           = \(\dfrac{98}{75}\)
=> \(\dfrac{1}{5}\)X+\(\dfrac{2}{5}\)+\(\dfrac{6}{15}\)X+\(\dfrac{31}{15}\)           = \(\dfrac{98}{75}\)
=> (\(\dfrac{1}{5}\)X+\(\dfrac{6}{15}\)X)+(\(\dfrac{2}{5}\)+\(\dfrac{31}{15}\))      =\(\dfrac{98}{75}\)
=> X x(\(\dfrac{1}{5}\)+\(\dfrac{6}{15}\))+(\(\dfrac{6}{15}\)+\(\dfrac{31}{15}\))    =\(\dfrac{98}{75}\)
=> X x(\(\dfrac{3}{15}\)+\(\dfrac{6}{15}\))+\(\dfrac{37}{15}\)            = \(\dfrac{98}{75}\)
=>X x\(\dfrac{9}{15}\)+\(\dfrac{37}{15}\)                      =\(\dfrac{98}{75}\)
=>X x\(\dfrac{9}{15}\)                             =\(\dfrac{98}{75}\)-\(\dfrac{37}{15}\)
=>X                                     =\(\dfrac{-29}{25}\):\(\dfrac{9}{15}\)
=>X                                     =\(\dfrac{-29}{15}\)

b) \(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)-2x=\dfrac{1}{2}\)

\(\Leftrightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}+.......-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)-2x=\dfrac{1}{2}\)

\(\Leftrightarrow\left(1-\dfrac{1}{100}\right)-2x=\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{99}{100}-2x=\dfrac{1}{2}\)

\(\Leftrightarrow-2x=\dfrac{1}{2}-\dfrac{99}{100}\)

\(\Leftrightarrow-2x=\dfrac{-49}{100}\)

\(\Leftrightarrow x=\dfrac{-49}{100}:\left(-2\right)\)

\(\Leftrightarrow x=\dfrac{49}{200}\)

Vậy ...............................

a) x = 2 hoặc 1

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