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.

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

chịu

A=1x3x(5+1) + 3x5x(7-1) +5x7x(9-3) +...+ 99x101x(103-97)

6A=3+ 1x3x5 +3x5x7-1x3x5 + 5x7x9 -3x5x7 +....+99x101x103 - 97x99x101

6A=3+99x101x103=1019703

vậy = 1019703

nếu sai chỗ nào thì sửa hộ mk vs

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)

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

\(=\frac{1}{2}.\frac{100}{101}\)

\(=\frac{50}{101}\)

\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{99\cdot101}\)

\(=2\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{99\cdot101}\right)\)

\(=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{99\cdot101}\)

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

\(=\frac{1}{1}-\frac{1}{101}=\frac{101}{101}-\frac{1}{101}=\frac{100}{101}\)

=1(2+1)+2(3+1)+3(4+1)+...+100(101+1)

=1.2+1+2.3+2+3.4+3+...+100.101+100

=(1.2+2.3+3.4+..+100.101)+(1+2+3+...+100)

=333300+5000

=338300

A=1x3 +3x5 +5x7 +....+99x101

6A=1x3x(5+1) + 3x5x(7-1) +5x7x(9-3) +...+ 99x101x(103-97)

6A=3+ 1x3x5 +3x5x7-1x3x5 + 5x7x9 -3x5x7 +....+99x101x103 - 97x99x101

6A=3+99x101x103=1019703