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C
11 tháng 1 2020
\(x-\frac{20}{11\cdot13}-\frac{20}{13\cdot15}-\frac{20}{15\cdot17}-......-\frac{20}{53\cdot55}=\frac{3}{11}\)
\(\Leftrightarrow x-10\left(\frac{2}{11\cdot13}-\frac{2}{13\cdot15}-\frac{2}{15\cdot17}-.....-\frac{2}{53\cdot55}\right)=\frac{3}{11}\)
\(\Leftrightarrow x-10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+....+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Leftrightarrow x-10\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Rightarrow x=1\)
G6
1
29 tháng 7 2018
\(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)
\(\Rightarrow x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)
\(\Rightarrow x-\left[10\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{20}{53.55}\right)\right]=\frac{3}{11}\)
\(\Rightarrow x-\left[10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)\right]=\frac{3}{11}\)
\(\Rightarrow x-\left[10\left(\frac{1}{11}-\frac{1}{55}\right)\right]=\frac{3}{11}\)
\(\Rightarrow x-\left[10.\frac{4}{55}\right]=\frac{3}{11}\)
\(\Rightarrow x-\frac{8}{11}=\frac{3}{11}\)
\(\Rightarrow x=\frac{3}{11}+\frac{8}{11}\)
\(\Rightarrow x=1\)
Vậy x = 1
_Chúc bạn học tốt_
G6
2
18 tháng 8 2018
\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+....+\frac{20}{53.55}\right)=\frac{3}{11}\)
\(x-10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10.\frac{4}{55}=\frac{3}{11}\)
\(x-\frac{8}{11}=\frac{3}{11}\Rightarrow x=\frac{3}{11}+\frac{8}{11}=\frac{11}{11}=1\)
18 tháng 8 2018
\(x-\left(\frac{20}{11\cdot13}+\frac{20}{13\cdot15}+...+\frac{20}{53\cdot55}\right)=\frac{3}{11}\)
Đặt \(x-A=\frac{3}{11}\)
\(\frac{A}{10}=\frac{2}{11\cdot13}+\frac{2}{13\cdot15}+...+\frac{2}{53\cdot55}\)
\(\frac{A}{10}=\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\)
\(\frac{A}{10}=\frac{1}{11}-\frac{1}{55}\)
\(\frac{A}{10}=\frac{4}{55}\)
\(A=\frac{8}{11}\)
\(\Rightarrow x-\frac{8}{11}=\frac{3}{11}\)
\(\Rightarrow x=\frac{3}{11}+\frac{8}{11}\)
\(\Rightarrow x=1\)
HT
28 tháng 6 2015
\(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-.....-\frac{20}{53.55}=\frac{3}{11}\)
\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+....+\frac{20}{53.55}\right)=\frac{3}{11}\)
\(x-10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+.....+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-\frac{40}{55}=\frac{3}{11}\)
\(x=\frac{3}{11}+\frac{40}{55}\)
\(x=1\)
7 tháng 8 2018
Bài 1:
a) \(x-\frac{20}{11.13}-\frac{20}{13.15}-...-\frac{20}{53.55}=\frac{3}{11}\)
\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)
\(x-\frac{20}{2}.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10.\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10\cdot\frac{4}{55}=\frac{3}{11}\)
\(x-\frac{8}{11}=\frac{3}{11}\)
\(x=1\)
b) \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(2.\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(2.\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{18}\)
=> x + 1 =18
x = 17
bài 2 ko bk lm, xl nha
TQ
0
3 tháng 7 2017
a ) Ta có : \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{5}\)
\(\Leftrightarrow\left(\frac{x+11}{10}-1\right)+\left(\frac{x+21}{10}-1\right)+\left(\frac{x+31}{30}-1\right)=\left(\frac{x+41}{40}-1\right)+\left(\frac{x+101}{50}-2\right)\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}=\frac{x+1}{40}+\frac{x+1}{50}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}-\frac{x+1}{40}-\frac{x+1}{50}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)\ne0\)
Nên x + 1 = 0
=> x = -1
28 tháng 6 2018
Đáp án: thiếu đề
@#@
mời bn xem xét lại đề bài.
~hok tốt~
x-20/11.13-20/13.15-...-20/53.55=3/11
=>x-(20/11.13+20/13.15+...+20/53.55)=3/11
Rồi tính tổng trong ngoặc là OK
ta có: x-\(\left(\frac{20}{11.13}-\frac{20}{13.15}-...-\frac{20}{53.55}\right)=\frac{3}{11}\)
\(x-\left[20\left(\frac{1}{11.13}-\frac{1}{13.15}-...-\frac{1}{53.55}\right)\right]=\frac{3}{11}\)
\(x-\left[20\left(\frac{2}{11.13}-\frac{2}{13.15}-...-\frac{2}{53.55}\right)\right]=\frac{3}{11}\)
\(x-\left[20\left(\frac{1}{11}+\frac{1}{13}-\frac{1}{13}+...-\frac{1}{53}+\frac{1}{55}\right)\right]=\frac{3}{11}\)
\(x-\left[20\left(\frac{1}{11}+\frac{1}{55}\right)\right]=\frac{3}{11}\)
\(x-\left(\frac{20.6}{55}\right)=\frac{3}{11}\)
\(x-\frac{24}{11}=\frac{3}{11}\)
\(x=\frac{3}{11}+\frac{24}{11}\)
\(x=\frac{27}{11}\)