Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}\)

\(< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)

\(B=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\) \(\Rightarrow A< \dfrac{99}{100}\)

\(1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-...-\dfrac{1}{100^2}=1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}\right)=1-A>\dfrac{1}{100}\)

\(\left(\dfrac{1}{2^2}-1\right)\times\left(\dfrac{1}{3^2-1}\right)\times\left(\dfrac{1}{4^2}-1\right)\times...\times\left(\dfrac{1}{100^2}-1\right)\)

\(=\dfrac{3}{2^2}\times\dfrac{8}{3^2}\times\dfrac{15}{4^2}\times...\times\dfrac{100^2-1}{100^2}\)

\(=\dfrac{1\times3}{2\times2}\times\dfrac{2\times4}{3\times3}\times\dfrac{3\times5}{4\times4}\times...\times\dfrac{99\times101}{100\times100}\)

\(=\dfrac{1\times2\times3\times...\times99}{2\times3\times4\times...\times100}\times\dfrac{3\times4\times5\times...\times101}{2\times3\times4\times...\times100}\)

\(=\dfrac{1}{100}\times\dfrac{101}{2}\)

\(=\dfrac{101}{200}\)

\(\left(\dfrac{1}{2^2}-1\right)\cdot\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)

\(=\dfrac{-3}{4}\cdot\dfrac{-8}{3}\cdot...\cdot\dfrac{-9999}{10000}\)

\(=\dfrac{1\cdot\left(-3\right)}{2\cdot2}\cdot\dfrac{2\cdot\left(-4\right)}{3\cdot3}\cdot...\cdot\dfrac{99\cdot\left(-101\right)}{100\cdot100}\)

\(=\dfrac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\cdot\dfrac{\left(-3\right)\cdot\left(-4\right)\cdot...\cdot\left(-101\right)}{2\cdot3\cdot4\cdot...\cdot100}\)

Ở tử số phân số bên phải có số thừa số là: \(101-3+1=99\)

99 là số lẻ nên tử số vế phải sẽ cho ra số âm.

\(=\dfrac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\cdot\dfrac{3\cdot4\cdot5\cdot...\cdot\left(-101\right)}{2\cdot3\cdot4\cdot...\cdot100}\)

\(=\dfrac{1\cdot\left(-101\right)}{100\cdot2}\)

\(=\dfrac{-101}{200}\)

Cảm ơn bạn nhiều ạ<3

Ta có:

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

\(\Rightarrow100-1-\dfrac{1}{2}-...-\dfrac{1}{100}=\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{99}{100}\)

\(\Rightarrow100=1+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{3}+...+\dfrac{1}{100}+\dfrac{99}{100}\)

\(\Rightarrow100=1+1+1+...+1\) (\(100\) số \(1\))

\(\Rightarrow100=100\)

Vậy \(100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)=\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{99}{100}\) (Đpcm)

Nguyễn Trần Thành ĐạtXuân Tuấn TrịnhHung nguyenHoang HungQuan Ace Legona giúp với

Ta có :

\(100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...............+\dfrac{1}{100}\right)\)

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

\(=99-\dfrac{1}{2}-\dfrac{1}{3}-................-\dfrac{1}{100}\)

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

\(=\dfrac{1}{2}+\dfrac{2}{3}+.................+\dfrac{99}{100}\)

Vậy :\(100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...............+\dfrac{1}{100}\right)=\dfrac{1}{2}+\dfrac{2}{3}+....................+\dfrac{99}{100}\)

\(\Rightarrowđpcm\)

a, \(\dfrac{x-1}{21}\) = \(\dfrac{3}{x+1}\)

   ( x-1)(x+1) = 21.3

    x² + x - x -1 = 63

     x²                = 63 + 1

     x²               = 64

    x = + - 8

b, 2\(\dfrac{1}{2}\)x + x = 2\(\dfrac{1}{17}\)

        x( \(\dfrac{5}{2}\) + 1) = \(\dfrac{35}{17}\)

       x              = \(\dfrac{35}{17}\) : ( \(\dfrac{5}{2}\)+1)

       x             = \(\dfrac{35}{17}\) x \(\dfrac{2}{7}\)

       x            = \(\dfrac{10}{17}\)

c, (x + \(\dfrac{1}{4}\) - \(\dfrac{2}{3}\) ) : ( 2 + \(\dfrac{1}{6}\) - \(\dfrac{1}{4}\)) = \(\dfrac{7}{46}\)

   (x  - \(\dfrac{5}{12}\)):  \(\dfrac{23}{12}\)                     =   \(\dfrac{7}{46}\)

  (x - \(\dfrac{5}{12}\))                               =   \(\dfrac{7}{46}\) x \(\dfrac{23}{12}\)

  x   - \(\dfrac{5}{12}\)                                =    \(\dfrac{7}{12}\)

 x                                            =    \(\dfrac{7}{12}\) + \(\dfrac{5}{12}\)

x                                             =     1

d, 2\(\dfrac{1}{3}\)x - 1\(\dfrac{3}{4}\)x + \(2\dfrac{2}{3}\)  = 3\(\dfrac{3}{5}\)

   x( \(\dfrac{7}{3}\) - \(\dfrac{7}{4}\)) + \(\dfrac{8}{3}\)      =  \(\dfrac{18}{5}\)

   x\(\dfrac{7}{12}\)                    = \(\dfrac{18}{5}\) - \(\dfrac{8}{3}\)

   x\(\dfrac{7}{12}\)                   = \(\dfrac{14}{15}\)

  x                         = \(\dfrac{14}{15}\) : \(\dfrac{7}{12}\)

 x                          = \(\dfrac{8}{5}\)

Lời giải:

Gọi phân số vế trái là $A$. Gọi tử số là $T$. Xét mẫu số:
\(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)

\(=1-\frac{1}{2}+1-\frac{1}{3}+1-\frac{1}{4}+....+1-\frac{1}{100}\)

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

\(=100-(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100})\)

\(=\frac{1}{2}\left[200-(3+\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100})\right]=\frac{1}{2}T\)

$\Rightarrow A=\frac{T}{\frac{1}{2}T}=2$ 

Ta có đpcm.

Giải:

Vì \(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}=2\) nên phần tử gấp 2 lần phần mẫu

Ta có:

\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)  

\(=\dfrac{2.\left[100-\left(\dfrac{3}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{100}\right)\right]}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\)

\(=\dfrac{2.\left[\left(2-\dfrac{3}{2}\right)+\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{1}{5}\right)+...+\left(1-\dfrac{1}{100}\right)\right]}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\) 

\(=\dfrac{2.\left(\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{4}{5}+...+\dfrac{99}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\) 

\(=2\) 

Vậy \(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}=2\left(đpcm\right)\) 

Chúc bạn học tốt!