Ta có: \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2019\cdot2020}+\dfrac{1}{2020\cdot2021}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2019}-\dfrac{1}{2020}+\dfrac{1}{2020}-\dfrac{1}{2021}\)

\(=\dfrac{1}{1}-\dfrac{1}{2021}=\dfrac{2021}{2021}-\dfrac{1}{2021}\)

\(=\dfrac{2020}{2021}\)

mà \(\dfrac{2020}{2021}< \dfrac{2021}{2021}=1\)

nên A<1

Ta có: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2020\cdot2021}+\dfrac{1}{2021\cdot2022}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}\)

\(=1-\dfrac{1}{2022}=\dfrac{2021}{2022}\)

1/1x2+1/2x3+1/3x4+...+1/2020x2021+1/2021x2022

=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2020-1/2021+1/2021-1/2022.

=1/1-1/2022

=2021/2022

\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\right)=2023x\)

\(\Rightarrow2022x+\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...-\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}+\dfrac{1}{2022}-\dfrac{1}{2023}\right)=2023x\)\(\Rightarrow2022x-2023x=-\left(1-\dfrac{1}{2023}\right)\)

\(\Rightarrow-x=-\dfrac{2022}{2023}\Leftrightarrow x=\dfrac{2022}{2023}\)

(x + 1/1.2) + (x + 1/2.3) + (x + 1/3.4) + ... + (x + 1/2022.2023) = 2023x

x + x + x + ... + x + 1/1.2 + 1/2.3 + ... + 1/2022.2023 = 2023x

2022x + 1 - 1/2 + 1/2 - 1/3 + ... + 1/2022 - 2023 = 2023x

2023x - 2022x = 1 - 1/2023

x = 2022/2023

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

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

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

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

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

=>x+1=11

=>x=10

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

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

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

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

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

=>\(x+1=3\)

=>\(x=2\)

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

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

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

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

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

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}=201\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=201\)

\(1-\frac{1}{x+1}=201\)

\(\frac{1}{x+1}=1-201\)

\(\frac{1}{x+1}=-200\)

\(\Rightarrow x+1=-\frac{1}{200}\)

\(x=-\frac{1}{200}-1\)

\(x=-\frac{201}{200}\)

Vậy \(x=-\frac{201}{200}\)

Gọi A = \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{x.\left(x+1\right)}=\dfrac{19}{20}\)

\(\Rightarrow\) A = \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\)

\(\Rightarrow\) A = 1 - \(\dfrac{1}{x+1}\)

\(\Rightarrow\) 1 - \(\dfrac{1}{x+1}\) = \(\dfrac{19}{20}\)

\(\Rightarrow1-\dfrac{19}{20}=\dfrac{1}{x+1}\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{20}\)

\(\Rightarrow\) x + 1 = 20\(\Rightarrow\) x=19