\(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)

\(\Rightarrow x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)

Đặt S = \(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+....+\frac{20}{53.55}\)

\(=10\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{53.55}\right)\)

\(=10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)\)

\(=10\left(\frac{1}{11}-\frac{1}{55}\right)\)

\(=10\cdot\frac{4}{55}=\frac{8}{11}\)

Thay A vào đề bài,ta được:

\(x-\frac{8}{11}=\frac{3}{11}\)

\(x=\frac{3}{11}+\frac{8}{11}\)

\(x=1\)

Đổi thay A thành thay S nha bạn

Ta có : 

\(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{1}\)

\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{1}\)

Đặt \(A=\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\)

\(A=10\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{53.55}\right)\)

\(A=10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)\)

\(A=10\left(\frac{1}{11}-\frac{1}{55}\right)\)

\(A=10.\frac{4}{55}=\frac{40}{55}=\frac{8}{11}\)

\(\Rightarrow x-\frac{8}{11}=\frac{3}{1}=3\)

\(x=3+\frac{8}{11}=\frac{33}{11}+\frac{8}{11}=\frac{41}{11}\)

Ủng hộ mk nha !!! ^_^

><in lỗi bn, 3/11 chứ k phải \(\frac{3}{1}\)

Ta có \(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)

\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)

\(\text{Đ}\text{ặt}:A=\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\)

\(A=\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\)

\(A=10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)\)

\(A=10\left(\frac{1}{11}-\frac{1}{55}\right)\)

\(A=10\left(\frac{1}{11}-\frac{1}{55}\right)\)

\(\Rightarrow x-\frac{8}{11}=\frac{3}{11}\)

\(=>x=1\)

Tự làm dễ mà 

Bài 3 : 

b) Ta có 1+ 2 + 3 +4 + ...+ x =15

Nên \(\frac{x\left(x+1\right)}{2}=15\)

\(x\left(x+1\right)=30\)

=> \(x\left(x+1\right)=5.6\)

=> x = 5

Bài 2:

h; \(\dfrac{2}{3}\)\(x\)  + 50% + \(x\) = \(\dfrac{1}{10}\)

    \(\dfrac{2}{3}\)\(x\) + \(\dfrac{1}{2}\)  + \(x\) = \(\dfrac{1}{10}\)

    (\(\dfrac{2}{3}\)\(x\) + \(x\)) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)

     \(x\) \(\times\) (\(\dfrac{2}{3}\) + 1) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)

      \(x\) \(\times\) \(\dfrac{5}{3}\) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)

      \(x\) \(\times\) \(\dfrac{5}{3}\) = \(\dfrac{1}{10}\) - \(\dfrac{1}{2}\)

      \(x\) \(\times\) \(\dfrac{5}{3}\) = \(\dfrac{-2}{5}\)

      \(x\)         = \(\dfrac{-2}{5}\): \(\dfrac{5}{3}\)

      \(x\)         =   - \(\dfrac{6}{25}\) 

Lớp 5 chưa học số âm em nhé. 

Bài 3:

a;  (\(x\) + 2) + (\(x\) + 4) + (\(x+6\)) + ... + (\(x+100\)) = 6000

     \(x\) + 2 + \(x\) + 4 + ... + \(x\) + 2 + 4 + 6 + ... + 100 = 6000

    (\(x\) + \(x\) + \(x\) + ... + \(x\)) + (2 + 4 + ... + 100) = 6000

     Xét dãy số 2; 4; ...;100;

Đây là dãy số cách đều với khoảng cách là: 4 - 2 = 2

Số số hạng của dãy số trên là: (100 - 2) : 2 + 1 = 50 (số)

 Theo bài ra ta có:

   \(x\) \(\times\) 50 + (100 + 2) \(\times\) 50 : 2 = 6000

   \(x\) \(\times\) 50 + 102 x 50 : 2 = 6000

    \(x\) \(\times\) 50 + (102 : 2) x 50 = 6000

    \(x\) x 50 + 51 x 50 = 6000

   \(x\) \(\times\) 50 + 2550 = 6000

    \(x\) x 50 = 6000 - 2550

    \(x\) x 50 = 3450

    \(x\) x 50 = 3450

   \(x\)           = 3450 : 50

   \(x\)           = 69

=4x(\(\frac{1}{11x13}\)+\(\frac{1}{13x15}\)+.......+\(\frac{1}{99x101}\))

=4x(\(\frac{1}{11}\)-\(\frac{1}{13}\)+\(\frac{1}{13}\)-\(\frac{1}{15}\)+....+\(\frac{1}{99}\)-\(\frac{1}{101}\))

4x(\(\frac{1}{11}\)-\(\frac{1}{101}\))

=4x \(\frac{90}{1111}\)

=\(\frac{360}{1111}\)

\(\frac{4}{11\times13}+\frac{4}{13\times15}+\frac{4}{15\times17}+...+\frac{4}{99\times101}\)

\(=\frac{4}{11}-\frac{4}{13}+\frac{4}{13}-\frac{4}{15}+\frac{4}{15}-\frac{4}{17}+...+\frac{4}{99}-\frac{4}{101}\)

\(=\frac{4}{11}-\frac{4}{101}\)

\(=\frac{360}{1111}\)

Đặt biểu thức là A . Ta co 

\(A=\frac{4}{11.13}+\frac{4}{13.15}+....+\frac{4}{99.101}\)

\(\Rightarrow2\left(\frac{2}{11.13}+\frac{2}{13.15}+....+\frac{2}{99.101}\right)\)

\(\Rightarrow A=2\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+.....+\frac{1}{99}-\frac{1}{101}\right)\)

\(\Rightarrow A=2\left(\frac{1}{11}-\frac{1}{101}\right)\)

\(\Rightarrow A=\frac{2.90}{1111}=\frac{180}{1111}\)

\(\left(\frac{2}{11x13}+\frac{2}{13x15}+\frac{2}{15x17}+\frac{2}{17x19}+\frac{2}{19x21}\right)\)) x 462 - y = 19

\(=\left(\frac{13-11}{11x13}+\frac{15-13}{13x15}+\frac{17-15}{15x17}+\frac{19-17}{17x19}+\frac{21-19}{19x21}\right)\)x 462 - y = 19

\(=\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+\frac{1}{17}-\frac{1}{19}+\frac{1}{19}-\frac{1}{21}\)) x 462 - y = 19

\(=\left(\frac{1}{11}-\frac{1}{21}\right)\)x 462 - y = 19

\(=\left(\frac{21}{231}-\frac{11}{231}\right)\)x 462 - y = 19

\(=\frac{10}{231}\)x 462 - y = 19

\(=20\)- y = 19

                y = 20 - 19

                y = 1

Mình làm bài này rồi

Đúng đó

Bài này có trong violympic

thanks pro