Question

Tìm x, biết:1/1 x 3+1/3x5+1/5x7+...+1/x(x+2)=20/41

Answer 1

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+3\right)}=\frac{20}{41}\)

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

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

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

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

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

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

=> x + 2 = 41

=> x = 41 - 2

=> x = 39

Vẫy x = 39

Answer 2

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

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

=> \(1-\frac{1}{x+2}=\frac{40}{41}\)

=> \(\frac{1}{x+2}=\frac{1}{41}\)

=> x + 2 = 41

=> x = 39