Question

[Maco]

Tìm x

1/3 + 1/6 + 1/10 +.....+ 2/x.(x + 1) = 39/40

Giúp mik với

Answer 1

[Huy hoàng indonaca]

đơn giản

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{39}{40}\)

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x.\left(x+1\right)}=\frac{39}{40}\)

\(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{39}{40}\)

\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{39}{40}\)

\(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{39}{40}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{39}{40}:2\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{39}{80}\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{39}{80}\)

\(\frac{1}{x+1}=\frac{1}{80}\)

\(\Rightarrow x+1=80\)

\(\Rightarrow x=80-1=79\)

Answer 2

[Đức Phạm]

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x\left(x+1\right)}=\frac{39}{40}\)

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{39}{40}\)

\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x\left(x+1\right)}=\frac{39}{40}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{39}{40}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{39}{40}\div2=\frac{39}{80}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{39}{80}=\frac{1}{80}\)

\(\Leftrightarrow x+1=80\Rightarrow x=80-1=79\)

Vậy \(x=79\)