\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{2019}{2020}\)

\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}=\dfrac{2019}{2020}\)

\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+4}=\dfrac{2019}{2020}\)

\(\Leftrightarrow\dfrac{x+4-x-1}{\left(x+1\right)\left(x+4\right)}=\dfrac{2019}{2020}\)

\(\Leftrightarrow\dfrac{3}{\left(x+1\right)\left(x+4\right)}=\dfrac{2019}{2020}\)

\(\Leftrightarrow\left(x+1\right)\left(x+4\right).2019=6060\)

<=> x = - 0,208387929

P/s: Số lạ zậy?Đề sai ko

6060 nha bạn

Ta có: 1 + ( 1  + 2 ) + ( 1 + 2 + 3 ) + ... + ( 1 + 2 + 3 +...+ 2020) 

= ( 1 + 1 + 1 +... + 1 ) + (2 + 2 +...+ 2 ) + ( 3 + 3+...+ 3 ) + ...+ 2020

Có 2020 số 1 ; 2019 số 2 ; 2018 số 3 ;... ; 1 số 2020 

=  2020 x 1 + 2019 x 2 + 2018 x 3 + ... + 2020x 1

=> \(M=\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2020\right)}{1\times2020+2\times2019+...+2020\times1}\)

= \(\frac{1\times2020+2\times2019+...+2020\times1}{1\times2020+2\times2019+...+2020\times1}=1\)

\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+....+\dfrac{1}{\left(x+2020\right)\left(x+2021\right)=1}\)

\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+...+\dfrac{1}{x+2020}-\dfrac{1}{x+2021}=1\)

\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+2021}=1\)

\(\Leftrightarrow\dfrac{\left(x+2021\right)-\left(x+1\right)}{\left(x+1\right)\left(x+2021\right)}=1\)

\(\Leftrightarrow\dfrac{x+2021-x-1}{\left(x+1\right)\left(x+2021\right)}=1\)

\(\Leftrightarrow\dfrac{2020}{\left(x+1\right)\left(x+2021\right)}=1\)

\(\Leftrightarrow\left(x+1\right)\left(x+2021\right)=2020\)

\(\Leftrightarrow x^2+2021x+x+2021=2020\)

\(\Leftrightarrow x^2+2022x=-1\)

\(\Leftrightarrow x\left(x+2022\right)=-1\)

Đến đây bạn chia trường hợp để giaỉ ra nghiệm nguyên nhé

M = 0

\(M=\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2020^2}\right)\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\right)-\left(\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\right)\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\right)\)

\(M=\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2020^2}\right)\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\right)(1-1)\)

\(M=\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2020^2}\right)\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2020^2}\right).0\)

\(M=0\)

Vì số bị trừ và số trừ gồm hai tích đảo ngược nhau nên M=0

C

C

C

\(A=\left(2020\times2019+2019\times2018\right)\times\left(1+\dfrac{1}{2}:1\dfrac{1}{2}-1\dfrac{1}{3}\right)\)

\(A=\left[2019\times\left(2020+2018\right)\right]\times\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)\)

\(A=4038\times2019\times\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)\)

\(A=4038\times2019\times0\)

\(A=0\)

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{x\cdot\left(x+1\right)}=\frac{2020}{2021}\)    

\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2020}{2021}\)   

\\(1-\frac{1}{x+1}=\frac{2020}{2021}\)   

\(\frac{1}{x+1}=1-\frac{2020}{2021}\)   

\(\frac{1}{x+1}=\frac{1}{2021}\)   

\(\Rightarrow x+1=2021\)   

\(x=2021-1\)   

\(x=2020\)

đk: \(x\ne\left\{0;-1\right\}\)

Ta có: \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{2020}{2021}\)

\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2020}{2021}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{2020}{2021}\)

\(\Leftrightarrow\frac{x}{x+1}=\frac{2020}{2021}\)

\(\Leftrightarrow2021x=2020x+2020\)

\(\Rightarrow x=2020\)

\(\dfrac{2}{5}\times\dfrac{1}{7}+\dfrac{2}{7}\times\dfrac{2}{5}\)

\(=\dfrac{2}{5}\times\left(\dfrac{1}{7}+\dfrac{2}{7}\right)\)

\(=\dfrac{2}{5}\times\dfrac{3}{7}\)

\(=\dfrac{6}{35}\)

\(x+\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}\)

\(x+\dfrac{1}{6}=\dfrac{3}{4}\)

\(x=\dfrac{9}{12}-\dfrac{2}{12}\)

\(x=\dfrac{7}{12}\)

\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2020}\right)+x=\dfrac{1}{2}\)

\(\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2019}{2020}+x=\dfrac{1}{2}\)

\(\dfrac{1}{2020}+x=\dfrac{1}{2}\)

\(x=\dfrac{1}{2}-\dfrac{1}{2020}\)

\(x=\dfrac{1010}{2020}-\dfrac{1}{2020}\)

\(x=\dfrac{1009}{2020}\)

\(\dfrac{2}{5}\times\dfrac{1}{7}+\dfrac{2}{7}\times\dfrac{2}{5}\)

\(=\dfrac{2}{5}\times\left(\dfrac{1}{7}+\dfrac{2}{7}\right)\)

\(=\dfrac{2}{5}\times\dfrac{3}{7}\)

\(=\dfrac{6}{35}\)

\(x+\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}\)

\(\Rightarrow\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}-x\)

\(\Rightarrow\dfrac{3}{4}-x=\dfrac{1}{6}\)

\(\Rightarrow x=\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{7}{12}\)

\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2020}\right)+x=\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2019}{2020}+x=\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1\times2\times3\times4\times...\times2019}{2\times3\times4\times5\times...\times2020}+x=\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{2020}+x=\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{1}{2}-\dfrac{1}{2020}=\dfrac{1009}{2020}\)