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

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

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

=>\(\frac{1}{x+1}=\frac{1}{11}\)

=>x+1=11

=>x=10

Do mỗi số hạng ở vế trái nằm trong dấu giá trị tuyệt đối mà vế phải 100 là số dương nên x cũng phải dương.

Do x dương và trong mỗi dấu giá trị tuyệt đối đều dương nên ta lập được kết quả sau:

x+1/1.2+x+1/2.3+1/3.4+....+x+1/99.100=100x

Dãy trên có 99 số x nên:

99x+(1-1/2+1/2-1/3+1/3-1/4+....+1/99-1/100)=100x

1-1/100=x

x=99/100

Vậy x=99/100

Chúc em học tốt^^

123

Ta có: 1/1.2+1/2.3+1/3.4+...+1/x(x+1)=2/3

=> 1-1/2+1/2-1/3+1/3-1/4+...+1/x-1/x+1=2/3

=>1-1/x+1=2/3

=>1/x+1=1/3

=>3=x+1

=>x=2

Ta có\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{3}\)

=>\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2}{3}\)

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

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

=>\(\frac{1}{x+1}=\frac{1}{3}\)

=>\(x+1=3\)

=>\(x=2\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{3}\)

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

\(\Rightarrow1-\frac{1}{x-1}=\frac{2}{3}\)

\(\Rightarrow\frac{1}{3}=\frac{1}{x-1}\)

\(\Rightarrow x=3+1=4\)

Ta có :\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2009}{2010}\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{2010}\)

\(\Rightarrow1-\frac{1}{x+1}=\frac{2009}{2010}\)

\(\Rightarrow\frac{1}{x+1}=1-\frac{2009}{2010}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2010}\)

\(\Rightarrow x+1=2010\)

\(\Rightarrow x=2010-1\)

\(\Rightarrow x=2009\)

Vậy x = 2009

=> 1-1/2+1/2-1/3+1/3- 1/4 +... +1/x -1/x+1 = 2009/1020

=> 1 - 1/x+1=2009/2010

=> (x+1-1)/x+1=2009/2010

=> x/x+1=2009/2010

=>x=2009

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..........+\frac{1}{19.20}-\frac{x}{40}=\frac{3}{-10}\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-........-\frac{1}{20}-\frac{x}{40}=\frac{-3}{10}\)

\(\Rightarrow1-\frac{1}{20}-\frac{x}{40}=\frac{-3}{10}\)

\(\Rightarrow\frac{40}{40}-\frac{2}{40}-\frac{x}{40}=\frac{-12}{40}\)

\(\Rightarrow\frac{38}{40}-\frac{x}{40}=\frac{-12}{40}\)

\(\Rightarrow\frac{x}{40}=\frac{38}{40}-\frac{-12}{40}\)

\(\Rightarrow\frac{x}{40}=\frac{38}{40}+\frac{12}{40}\)

\(\Rightarrow\frac{x}{40}=\frac{50}{40}\)

\(\Rightarrow x=50\)

Vậy x = 50

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+..+\frac{1}{19\cdot20}-\frac{x}{40}=\frac{-3}{10}\)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{19}-\frac{1}{20}-\frac{x}{40}=\frac{3}{-10}\)

\(1-\frac{1}{20}-\frac{x}{40}=\frac{3}{-10}\)

\(\frac{x}{40}=1-\frac{1}{20}-\frac{3}{-10}=1\frac{1}{4}=\frac{5}{4}\)

\(\frac{x}{40}=\frac{5}{4}\Rightarrow x=\frac{40\cdot5}{4}=50\)

\(x\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ x\cdot\left(1-\dfrac{1}{50}\right)=1\\ \dfrac{49}{50}x=1\\ x=1:\dfrac{49}{50}\\ x=\dfrac{50}{49}\)