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!

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 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

giúp mình với

a; 1 + 2 + 3 + ... + \(x\) = 5050

   Số số hạng của dãy số trên là: (\(x-1\)):1 + 1 = \(x\)

    (\(x\) + 1)\(\times\) \(x\): 2 =  5050

    (\(x\) + 1) \(\times\) \(x\)   = 5050 \(\times\) 2

    (\(x+1\)) \(\times\) \(x\)  = 10100

   (\(x+1\)) \(\times\) \(x\) = 101 \(\times\) 100

   Vậy \(x=100\)

b;  \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + ... + \(\dfrac{1}{x^2+x}\) = \(\dfrac{99}{100}\)

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

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

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

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

              \(x+1\) = 100

              \(x=100-1\)

              \(x=99\)

         Vậy \(x=99\)

a)

Ta có : ( 1 + 2 + 3 + ... + 99)

Số số hạng là:       ( 99 - 1 )  : 1 + 1 = 100

Tổng là:                 ( 99 + 1 ) x 100 : 2 = 5000

=> 5000 x ( 13  - 12 - 1 ) x 15

=> 5000 x 10 x 15

=> 50000 x 15

=> 750000

Ko muốn vt nx :))

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

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

1/1x2+1/2x3+1/3x4+...+1/x=99?100

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

1/1-1/x=99/100

1/x=1/1-99/100

1/x=1/100

=>x=100

kbn nha

Nhấn vào đây : 

Câu hỏi của Nguyễn Thanh Tùng - Toán lớp 6 - Học toán với OnlineMath