Question

1/1.3 + 1/3.5 + 1/5.7 +...+ 1/x.(x+2) = 20/41

Answer 1

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

Answer 2

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

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

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

=> \(1-\frac{1}{x+2}=\frac{40}{41}\)

=> \(\frac{1}{x+2}=1-\frac{40}{41}\)

=> \(\frac{1}{x+2}=\frac{1}{41}\)

=> \(x+2=41\)

=> \(x=41-2=39\)

Answer 3

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

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

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

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

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

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

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

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

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