Question

Tìm X ( x này là nhân )

_{\(\frac{1}{1x2}+\frac{1}{2x3}+.....+\frac{1}{Xx\left(x+1\right)}=\frac{9}{10}\)}

Answer 1

Sửa đề:

\(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{x\times\left(x+1\right)}=\frac{9}{10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{9}{10}\)

\(1-\frac{1}{x}=\frac{9}{10}\)

\(\frac{1}{x}=1-\frac{9}{10}=\frac{1}{10}\)

Vậy, x = 10.

Ko bt có right ko?

Answer 2

Nhầm.

Chuyển \(1-\frac{1}{x}\)thành \(1-\frac{1}{x+1}\)

\(1-\frac{1}{x+1}=\frac{9}{10}\)

\(\frac{1}{x+1}=1-\frac{9}{10}=\frac{1}{10}\)

Vậy x = 10 - 1 = 9

Thế ms right chứ!

Answer 3

\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{x\left(x+1\right)}=\frac{9}{10}\)

\(\Leftrightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{9}{10}\)

\(\Leftrightarrow\frac{1}{1}-\frac{1}{x+1}=\frac{9}{10}\Leftrightarrow\frac{1}{x+1}=\frac{1}{1}-\frac{9}{10}=\frac{1}{10}\)

\(\Leftrightarrow x+1=10\Rightarrow x=10-1=9\)

Vậy x = 9