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!!!

\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\right)\cdot100-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ \left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\cdot100-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ \left(1-\frac{1}{10}\right)\cdot100-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ \frac{9}{10}\cdot100-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ 90-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ \left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=1\\ \frac{5}{2}:\left(X+\frac{206}{100}\right)=\frac{1}{2}\\ X+\frac{206}{100}=5\\ X=\frac{500}{100}-\frac{206}{100}\\ X=\frac{294}{100}=\frac{147}{50}\)

Vậy \(X=\frac{147}{50}\)

( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ......+ 1/9 - 1/10) . 100 - [ 5/2 : ( x + 103/50 ) ] = 89 . 1/2

( 1 - 1/10) . 100 - [ 5/2 : ( x + 103/50 ) ] = 89/2

90 - 5/2 : ( x + 103/50 ) = 89/2

5/2 : ( x + 103/50 ) = 90 - 89/2

5/2 : ( x + 103/50 ) = 91/2

x + 103/50 = 5/2 : 91/2

x + 103/50 = 5/91

x = 5/91 - 103/50

x = -9,123/4550

\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}=\frac{2-1}{1.2}+\frac{4-3}{3.4}+\frac{6-5}{5.6}+...+\frac{100-99}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}=\left(\frac{1}{1}+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=\left(\frac{1}{1}+\frac{1}{3}+...+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

=> \(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}=\frac{1}{1.2}+\frac{1}{3.4}+....+\frac{1}{99.100}\)

=> A = 1

A=1 là đúng rồi

a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

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

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

\(\Rightarrow A< 1\)

b) \(B=\frac{1}{3}+\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{3}\right)^{100}\)

\(\Rightarrow3B=1+\frac{1}{3}+...+\left(\frac{1}{3}\right)^{99}\)

\(\Rightarrow3B-B=1-\left(\frac{1}{3}\right)^{100}\)

\(\Rightarrow2B=1-\left(\frac{1}{3}\right)^{100}< 1\)

\(\Rightarrow2B< 1\)

\(\Rightarrow B< \frac{1}{2}\)

c)\(C=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-1-\frac{1}{2}-...-\frac{1}{25}\)

\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)

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

\(\Leftrightarrow3x+\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right)=1\)

\(\Leftrightarrow3x+\frac{3}{2}=1\)

\(\Leftrightarrow3x=-\frac{1}{2}\)

\(\Leftrightarrow x=-\frac{1}{2}\div3=-\frac{1}{6}\)

Sửa đề \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{x.\left(x+1\right)}=\frac{99}{100}\)

\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2}-\frac{1}{x+1}=\frac{99}{100}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{99}{100}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{100}\)

\(\Leftrightarrow x=99\)

a) => ( x + 1/2 ) . 3 = 1

=> 3x + 3/2 = 1

=> 3x = 1 - 3/2

=> 3x = -1/2

=> x = -1/2 : 3 = -1/6

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

\(\Leftrightarrow3\left(x+\frac{1}{2}\right)=1\)

\(\Leftrightarrow x+\frac{1}{2}=\frac{1}{3}\)

\(\Leftrightarrow x=\frac{1}{3}-\frac{1}{2}\)

\(\Leftrightarrow x=-\frac{1}{6}\)