Ta có: \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{20}{41}\)

\(\Leftrightarrow\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{40}{41}\)

\(\Leftrightarrow1-\dfrac{2}{x+2}=\dfrac{40}{41}\)

\(\Leftrightarrow\dfrac{2}{x+2}=\dfrac{1}{41}\)

Suy ra: x+2=82

hay x=80

biểu thức trên nó bằng bao nhiêu hả bạn

bằng 20/41 quên ghi

Bieu thức đó bằng mấy đã bn! Biểu thức đó phải có giá trị mới tìm được x chứ bn!

Tk mình đi nhé!

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)
\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}\right)=\frac{20}{41}\)
\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\frac{1}{2}.\frac{x+1}{x+2}=\frac{20}{41}\)
\(\frac{x+1}{x+2}=\frac{20}{41}:\frac{1}{2}\)
\(\frac{x+1}{x+2}=\frac{40}{41}\)
\(x+1=40 \)
\(x=40-1\)
\(x=39\)
Đúng thì ****

Lương Hồ Khánh Duy trả lời đúng nhưng đúng cảu bài khác

Ở đây, câu hỏi ghi x+1 bn ghi x+2

Trần Quang Trường, Lương Hồ Khánh Duy đã trả lời đúng rồi , nếu câu hỏi như bạn nói thì phép tính ko như quy luật của nó 

Ta có:

1/1.3 + 1/3.5 + 1/5.7 + ... + 1/x.(x+2) = 1/2.(2/1.3 + 2/3.5 + 2/5.7 + ... + 2/x.(x+2)

                                                          = 1/2.(1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/x - 1/x+2

                                                          = 1/2.(1 - 1/x+2)

=> 1/2.(1 - 1/x+2) = 20/41

            1 - 1/x+ 2  = 20/41 : 1/2

            1 - 1/x+2   = 40/41

                  1/x+2  = 1/41

=>x + 2 = 41

=>x        = 41 - 2

=>x        = 39

Vậy x = 39

Ủng hộ nha

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)

=> \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}=2.\frac{20}{41}\)

=> \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{40}{41}\)

=> \(1-\frac{1}{x+2}=\frac{40}{41}\)

=> \(\frac{1}{x+2}=1-\frac{40}{41}\)

=> \(\frac{1}{x+2}=\frac{1}{41}\)

=> \(x+2=41\)

=> \(x=41-2=39\)

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)

\(\Leftrightarrow\)\(\frac{2}{1.3}+\frac{2}{3.4}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}=\frac{40}{41}\)

\(\Leftrightarrow\)\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{40}{41}\)

\(\Leftrightarrow\)\(1-\frac{1}{x+2}=\frac{40}{41}\)

\(\Leftrightarrow\)\(\frac{1}{x+2}=1-\frac{40}{41}\)

\(\Leftrightarrow\)\(\frac{1}{x+2}=\frac{1}{41}\)

\(\Leftrightarrow\)\(x+2=41\)

\(\Leftrightarrow\)\(x=41-2\)

\(\Leftrightarrow\)\(x=39\)

Đề sai rồi bạn ơi

Phải là 1/x.(x+2)

Gọi tổng trên là A

1/2A= 2/1.3+1/3.5+...+1/x.(x+2)

1/2A= 1-1/x.(x+2)

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

cảm ơn

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)

\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{20}{41}\)

\(\Leftrightarrow1-\frac{1}{x+2}=\frac{21}{41}\)

\(\Leftrightarrow\frac{1}{x+2}=1-\frac{21}{41}\)

\(\Leftrightarrow\frac{1}{x+2}=\frac{20}{41}\)

\(\Leftrightarrow20\left(x+2\right)=41\)

\(\Leftrightarrow x-2=\frac{41}{20}\)

\(\Leftrightarrow x=\frac{41}{20}+2\)

\(\Leftrightarrow x=\frac{81}{20}\)

\(\frac{1}{1.3}+...+\frac{1}{a\left(a+2\right)}=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{2}{a\left(a+2\right)}\right)=\frac{1}{2}\left(1-\frac{1}{3}+....-\frac{1}{a+2}\right)\) 

\(=\frac{1}{2}\left(1-\frac{1}{a+2}\right)=\frac{20}{41}\Rightarrow a+2=41\Leftrightarrow a=39\)

\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{x.\left(x+2\right)}\right)=\frac{20}{41}\)

\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)

\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)

\(\Rightarrow1-\frac{1}{x+2}=\frac{20}{41}:\frac{1}{2}\)

\(\Rightarrow1-\frac{1}{x+2}=\frac{40}{41}\)

\(\Rightarrow\frac{1}{x+2}=1-\frac{40}{41}\)

\(\Rightarrow\frac{1}{x+2}=\frac{1}{41}\)

\(\Rightarrow x+2=41\)

\(\Rightarrow x=41-2\)

\(\Rightarrow x=39\)

Vậy x = 39

bạn nào giải giúp mình với

nếu đúng thì mình sẽ ***

=1/2*(1-1/3+1/3-1/5+....+1/x+1/x+2)

=1/2*(1-1/x+2)

=>1/2*x+1/x+2=20/21

Đến đó đưa về giống tìm x nha

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)

\(=2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}\right)=\frac{20}{41}\)

\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}=\frac{40}{41}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{40}{41}\)

\(=1-\frac{1}{x+2}=\frac{40}{41}\)

\(\frac{1}{x+2}\)\(=1-\frac{40}{41}\)

\(\frac{1}{x+2}=\frac{1}{41}\)

=> x+2=41

          x=41-2

          x=39

a) Ta có 

1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/x - 1/x + 2 = 5

rút gọn ta được : 1 - 1/x+2 = 5

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

<=> x + 1 / x + 2 = 5

<=> x+ 1 = 5x + 10

<=> - 4x = 9

<=> x = -9/4

B) / x +4/ = 2^0 + 1 ^ 2013

=> /x + 4/ = 1 + 1

=> / x + 4 / = 2

TH 1 :  x+ 4 = 2

 => x = 2 - 4 = -2

TH2 : x + 4 = -2

=> x = -2 - 4 = -6

=> x = { - 2 , -6 }