\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x}=\frac{99}{100}\)

Đặt \(x=n.\left(n+1\right)\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n.\left(n+1\right)}=\frac{99}{100}\)

\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{n+1-n}{n.\left(n+1\right)}=\frac{99}{100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{\left(n+1\right)}=\frac{99}{100}\)

\(=1-\frac{1}{\left(n+1\right)}=\frac{99}{100}\)

\(\frac{1}{\left(n+1\right)}=1-\frac{99}{100}=\frac{1}{100}\)

\(\Rightarrow x=\left(100-1\right).100\)

         \(=9900\)

Ta có : \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+......+\frac{1}{x}=\frac{99}{100}\)

\(\Leftrightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{x}=\frac{99}{100}\)

\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{x}=\frac{99}{100}\)

Đề kì z 

thank you bạn nha

ta có

1+2+3+.........+x=5050

=>\(\frac{x.\left(x+1\right)}{2}=5050\)

=>x.(x+1)=5050.2

=>x.(x+1)=10100

=>x.(x+1)=100.101

=>x=100

giúp mình với ,cần gấp

Giải:

1/2 + 1/6 + 1/12 + ... + 1/x.(x+1) = 99/100

=>1/1.2 + 1/2.3 + 1/3.4 + ... + 1/x.(x+1) = 99/100

1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/x - 1/x+1 = 99/100

1/1 - 1/x+1 = 99/100

        1/x+1 = 1/1 - 99/100

        1/x+1 = 1/100

=>x+1 = 100

        x = 100 - 1

        x = 99

Vậy ...

Chúc bạn học tốt!

`1/2+1/6+1/12+....+1/(x(x+1))=99/100`

`-> 1/(1.2)+1/(2.3)+1/(3.4)+....+1/(x(x+1))=99/100`

`-> 1 - 1/2 + 1/2 -1/3 +1/3-1/4+....+1/x -1/(x+1)=99/100`

`-> 1-1/(x+1)=99/100`

`-> x/(x+1)=99/(99+1)`

`-> x=99`

Bài 1: 

ĐKXĐ: \(x\notin\left\{0;-1\right\}\)

Ta có: \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{99}{100}\)

\(\Leftrightarrow\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{99}{100}\)

\(\Leftrightarrow\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{99}{100}\)

\(\Leftrightarrow\dfrac{1}{1}-\dfrac{1}{x+1}=\dfrac{99}{100}\)

\(\Leftrightarrow\dfrac{x+1-1}{x+1}=\dfrac{99}{100}\)

\(\Leftrightarrow100x=99\left(x+1\right)\)

\(\Leftrightarrow100x-99x=99\)

hay x=99(thỏa ĐK)

Vậy: x=99

\(\Leftrightarrow\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{x.\left(x+1\right)}\)=\(\dfrac{99}{100}\)

\(\Leftrightarrow\)\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\)=\(\dfrac{99}{100}\)

\(\Leftrightarrow1-\dfrac{1}{x+1}\)=\(\dfrac{99}{100}\)

\(\Leftrightarrow\dfrac{1}{x+1}\)=\(\dfrac{1}{100}\)

\(\Leftrightarrow x+1\)=100

\(\Leftrightarrow\)\(x\)=99

ta gọi \(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\)là A

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(\Leftrightarrow1.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\right)\)

\(\Rightarrow A=1-\frac{1}{10}=\frac{9}{10}\)

ta gọi B là biểu thức thứ2

\(B=\frac{2.2}{3}\times\frac{3.3}{2.4}\times\frac{4.4}{3.5}\times...\times\frac{10.10}{9.11}\)

\(\Rightarrow\)2 x \(\frac{10}{11}\)\(=\frac{20}{11}\)

\(\Rightarrow\)\(x+\frac{9}{10}=\frac{20}{11}+\frac{9}{110}\)

\(\Rightarrow x=1\)

mk nghĩ vậy bạn ạ, mk mong nó đúng