Sửa đề: \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}=\frac{2020}{2021}\) \(Đkxđ:\hept{\begin{cases}x\ne0\\x\ne-2\end{cases}}\)

\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{2020}{2021}\)

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

\(\Leftrightarrow\frac{x+2}{2021}=1\)

\(\Leftrightarrow x=2019\)

Vậy \(x=2019\)

=>2/1*3+2/3*5+...+2/(2x-1)(2x+1)=98/99

=>1-1/3+1/3-1/5+...+1/(2x-1)-1/(2x+1)=98/99

=>1-1/(2x+1)=98/99

=>1/(2x+1)=1/99

=>2x+1=99

=>x=49

a)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}=\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+...+\left(\frac{1}{99}-\frac{1}{101}\right)\)

                                                               \(=1-\frac{1}{101}=\frac{100}{101}\)

b) \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}=\frac{2}{1.3}.\frac{5}{2}+\frac{2}{3.5}.\frac{5}{2}+\frac{2}{5.7}.\frac{5}{2}+...+\frac{2}{99.101}.\frac{5}{2}\)

                                                                \(=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)

                                                                \(=\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)

a)100/101

b)250/101

A)100/101

B)250/101

a.2/1.3+2/3.5+2/5.7+................+2/99.101

1-1/3+1/3-1/5+1/5-1/7+....+1/99-1/101

1-1/101

100/101

b.5/1.3+5/3.5+5/5.7+............+5/99.101

5.2/1.3.2+5.2/3.5.2+5.2/5.7.2+........+5.2+99.101.2

5/2(2/1.3+2/3.5+2/5.7+........+2/99.101)

5/2(1-1/3+1/3-1/5+1/5-1/7+........+1/99-1/101)

5/2(1-1/101)

5/2.100/101

250/101

Có:

\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{x.\left(x+2\right)}=\dfrac{5}{11}\)

\(\Rightarrow\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{5}{11}\)

\(\Rightarrow\dfrac{1}{2}.\left(1-0-0-0...-0-\dfrac{1}{x+2}\right)=\dfrac{5}{11}\)

\(\Rightarrow\dfrac{1}{2}.\left(1-\dfrac{1}{x+2}\right)=\dfrac{5}{11}\)

\(\Rightarrow1-\dfrac{1}{x+2}=\dfrac{5}{11}:\dfrac{1}{2}=\dfrac{10}{11}\)

\(\Rightarrow\dfrac{1}{x+2}=1-\dfrac{10}{11}\)

\(\Rightarrow\dfrac{1}{x+2}=\dfrac{1}{11}\)

\(\Rightarrow x+2=11\)

\(\Rightarrow x=11-2=9\)

Vậy x = 9.

Chúc bạn học tốt!

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

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

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

= 1/2.(1/1 - 0 - 1/x+2 )

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

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

= 1/2 . x+1/x+2

Mà 1/1.3 + 1/3.5 + 1/5.7 + ... +1/x.(x+2) = 5/11

=> 1/2 . x+1/x+2 = 5/11

=> x+1/x+2 = 5/11 : 1/2

=> x+1/x+2 = 10/11

=> x+1/x+2-1 = 10/11-1

=> x+1/x+1 = 10/10

=> x + 1 = 10

=> x = 10 - 1

=> x = 9

Vậy x = 9

\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{x.\left(x+2\right)}=\dfrac{5}{11}\)

\(\dfrac{1.2}{2.1.3}+\dfrac{1.2}{2.3.5}+\dfrac{1.2}{2.5.7}+...+\dfrac{1.2}{2x\left(x+2\right)}\dfrac{5}{11}\)\(\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{x\left(x+2\right)}\right)=\dfrac{5}{11}\)\(\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{5}{11}\)\(\dfrac{1}{2}\left(1+0+0+...+0+\dfrac{1}{x-2}\right)=\dfrac{5}{11}\)

\(\dfrac{1}{2}\left(1-\dfrac{1}{x-2}\right)=\dfrac{5}{11}\)

\(1-\dfrac{1}{x-2}=\dfrac{5}{11}:\dfrac{1}{2}\)

\(1-\dfrac{1}{x-2}=\dfrac{10}{11}\)

\(\dfrac{1}{x-2}=1-\dfrac{10}{11}\)

\(\dfrac{1}{x-2}=\dfrac{1}{11}\)

\(\Rightarrow x-2=11\)

\(x=11+2\)

\(x=13\)

Vậy x=13

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

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

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

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

=> \(1-\frac{1}{x+2}=\frac{40}{41}\)

=> \(\frac{1}{x+2}=1-\frac{40}{41}\)

=> \(\frac{1}{x+2}=\frac{1}{41}\)

=> \(x+2=41\)

=> \(x=41-2=39\)

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

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

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

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

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

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

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

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

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