a) \(\frac{1}{1.3}+\frac{1}{3.5}+....+\frac{1}{47.49}=\frac{24}{x+4}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+......+\frac{2}{47.49}\right)=\frac{24}{x+4}\)

\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{47}-\frac{1}{49}\right)=\frac{24}{x+4}\)

\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{49}\right)=\frac{24}{x+4}\)

\(\Rightarrow\frac{1}{2}.\frac{48}{49}=\frac{24}{x+4}\)

\(\Rightarrow\frac{24}{49}=\frac{24}{x+4}\)

\(\Rightarrow x+4=49\Rightarrow x=45\)

\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{47.49}=\frac{24}{x+4}\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{47.49}\right)=\frac{24}{x+4}\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{47}-\frac{1}{49}\right)=\frac{24}{x+4}\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{49}\right)=\frac{24}{x+4}\Leftrightarrow A=\frac{1}{2}.\frac{48}{49}=\frac{24}{x+4}\)

\(\Rightarrow A=\frac{24}{49}=\frac{24}{x+4}\Leftrightarrow x=49-4=45\)

Bài b) hình như sai đề thì phải đó bạn.

a)ta co \(\frac{1}{1.3}+\frac{1}{3.5}+..+\frac{1}{47.49}=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{47}-\frac{1}{49}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{49}\right)=\frac{24}{49}=\frac{24}{x+4}\)\(\Rightarrow49=x=4\Rightarrow x=45\)

b)ta co \(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{97.101}=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{101}\)

\(=1-\frac{1}{101}=\frac{100}{101}=\frac{2x+4}{101}\Rightarrow100=2x+4\Rightarrow x=48\)

Sửa đề: \(\dfrac{4}{1.3}+\dfrac{4}{3.5}+...+\dfrac{4}{99.101}\)

Đặt: \(A=\dfrac{4}{1.3}+\dfrac{4}{3.5}+...+\dfrac{4}{99.101}\)

\(=2\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}\right)\)

\(=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=2\left(1-\dfrac{1}{101}\right)=\dfrac{200}{101}\)

*Lưu ý: Dấu ".'' trong bài là dấu nhân nhé, lên lớp 6 bạn sẽ được học

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

Bạn ơi đề phải là vầy chứ nhỉ?

\(\dfrac{4}{1\times3}+\dfrac{4}{3\times5}+...+\dfrac{4}{99\times101}\)

\(=\dfrac{4}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=2.\left(1-\dfrac{1}{101}\right)\)

\(=2.\left(\dfrac{101}{101}-\dfrac{1}{101}\right)\)

\(=2.\dfrac{100}{101}\)

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

\(#NqHahh\)

ta có : 2S=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

          2S=\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

          2S=\(\frac{1}{1}-\frac{1}{101}\)

      2S+\(\frac{1}{101}\)= \(\frac{1}{1}-\frac{1}{101}+\frac{1}{101}\)

      2S+\(\frac{1}{101}\)=1

ok

Lời giải:
Xét thừa số tổng quát $1+\frac{1}{n(n+2)}=\frac{n(n+2)+1}{n(n+2)}=\frac{(n+1)^2}{n(n+2)}$

Khi đó:

$1+\frac{1}{1.3}=\frac{2^2}{1.3}$

$1+\frac{1}{2.4}=\frac{3^2}{2.4}$

.........

$1+\frac{1}{99.101}=\frac{100^2}{99.101}$

Khi đó:

$A=\frac{2^2.3^2.4^2......100^2}{(1.3).(2.4).(3.5)....(99.101)}$

$=\frac{(2.3.4...100)(2.3.4...100)}{(1.2.3...99)(3.4.5...101)}$

$=\frac{2.3.4...100}{1.2.3..99}.\frac{2.3.4...100}{3.4.5..101}$
$=100.\frac{2}{101}=\frac{200}{101}$

giúp em với

vcbdf

\(\left(1+\frac{1}{1\times3}\right)\times\left(1+\frac{1}{2\times4}\right)\times\left(1+\frac{1}{3\times5}\right)\times...\times\left(1+\frac{1}{99.101}\right)\)

\(=\left(\frac{3}{3}+\frac{1}{3}\right)\times\left(\frac{8}{8}+\frac{1}{8}\right)\times\left(\frac{15}{15}+\frac{1}{15}\right)\times...\times\left(\frac{9999}{9999}+\frac{1}{9999}\right)\)

\(=\frac{4}{3}\times\frac{9}{8}\times\frac{16}{15}\times...\times\frac{10000}{9999}\)

\(=\frac{4\times9\times16\times...\times10000}{3\times8\times15\times...\times9999}\)

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

\(=\frac{2\times100}{101}=\frac{200}{101}\)

mk cx co dap an vay

$\dfrac{1}{2+3}$