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\(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)
\(x-10\left(\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{53.55}\right)=\frac{3}{11}\)
\(x-10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10.\frac{4}{55}=\frac{3}{11}\)
\(x=1\)
-Nếu |x-2013|-2014=2015->|x-2013| = 4029
+ Nếu x-2013 =4029 -> x= 6042
+ Nếu x-2013 = -4029 -> x = -2016
- Nếu |x-2013|-2014= -2015 -> |x-2013| = -1 (loại vì |x-2013| \(\ge\)0 )
Vậy x= 6042 ; x=-2016 là các giá trị cần tìm.
\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+\frac{20}{15.17}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)
=> \(x-\left(20\times\frac{1}{11.13}+20\times\frac{1}{13.15}+20\times\frac{1}{15.17}+...+20\times\frac{1}{53.55}\right)=\frac{3}{11}\)
\(x-20\times\left(\frac{1}{11.13}+\frac{1}{13.15}+\frac{1}{15.17}+...+\frac{1}{53.55}\right)=\frac{3}{11}\)
\(x-20\times\left(\frac{1}{2}\times\left(\frac{1}{11}-\frac{1}{13}\right)+\frac{1}{2}\times\left(\frac{1}{13}-\frac{1}{15}\right)+\frac{1}{2}\times\times+...+\frac{1}{2}\times\left(\frac{1}{53}-\frac{1}{55}\right)\right)=\frac{3}{11}\)
\(x-20\times\frac{1}{2}\times\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-...-\frac{1}{53}+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10\times\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10\times\frac{4}{55}=\frac{3}{11}\)
\(x-\frac{10}{11}=\frac{3}{11}\)
=> \(x=\frac{3}{11}+\frac{10}{11}=\frac{13}{11}\)
Vậy x=\(\frac{13}{11}\)
a. nhân cả hai vế của đẳng thức với 1/ 10 ta có
x/10 - (2/11.13 +2/13.15+...+2/53.55)=3/11 . 1/10
x/10 - (1/11-1/13+1/13-1/15 +...+1/53-1/55) =3/110
x/10 - (1/11 - 1/55) =3/110
x/10 -4/55 = 3/110
x/10=3/110 + 4/55
x. 1/10 =1/10
x= 1/10 : 1/10 =1
b) bạn nhân cả hai vế của đẳng thức với 1/2 rồi làm tương tự
a. nhân cả hai vế của đẳng thức với \(\frac{1}{10}\). Ta có:
\(\frac{x}{10}-\left(\frac{2}{11.13}+\frac{2}{13.15}+...\frac{2}{53.55}\right)=\frac{3}{11}.\frac{1}{10}\)
\(\frac{x}{10}-\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{110}\)
\(\frac{x}{10}-\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{110}\)
\(\frac{x}{10}-\frac{-4}{55}=\frac{3}{110}\)
\(\frac{x}{10}=\frac{3}{110}+\frac{4}{55}\)
\(x.\frac{1}{10}=\frac{1}{10}\)
\(x=\frac{1}{10}:\frac{1}{10}=1\)
b. cũng thế bạn nhân hai vế của đẳng thức với \(\frac{1}{2}\) rồi làm tương tự.
x-10( 2/11.13+2/13.15+...+2/53.55)=3/11
x-10(1/11-1/55)=3/11
x-10.4/55=3/11
x-40/55=3/11
x=3/11+40/55
x= 1
x-10.(2/11.13+2/13.15+....+2/53.55)=3/11
x-10.(1/11-1/13+...+1/53-1/55)=3/11
x-10.(1/11-1/55)=3/11
x-10.4/55=3/11
x-8/11=3/11
x=1
Nhớ k cho mình nhé
Bài 1:
a)\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(=1-\frac{1}{2017}\)
\(=\frac{2016}{2017}\)
b)\(=1008\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(=1008\cdot\left(1-\frac{1}{2017}\right)\)
\(=1008\cdot\frac{2016}{2017}\)\(=\frac{290304}{31}\)Bài 2:
a)\(A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\)
\(=\frac{1}{3}-\frac{1}{21}\)
\(=\frac{2}{7}\)
b)\(B=\frac{5}{28}+\frac{5}{70}+...+\frac{5}{700}\)
\(=\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{25.28}\)
\(=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(=\frac{5}{3}\left(\frac{1}{4}-\frac{1}{28}\right)\)
\(=\frac{5}{3}\cdot\frac{6}{28}\)
\(=\frac{15}{14}\)
Bài 3:
a)Đặt \(A=-\frac{20}{11.13}-\frac{20}{13.15}-...-\frac{20}{53.55}\)
\(=-\left(\frac{20}{11.13}+\frac{20}{13.15}+...+\frac{20}{53.55}\right)\)
\(=-\left[10\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)\right]\)
\(=-\left[10\left(\frac{1}{11}-\frac{1}{55}\right)\right]\)
\(=-\left[10\cdot\frac{4}{55}\right]\)
\(=-\frac{8}{11}\).Thay vào ta có: \(x-\frac{8}{11}=\frac{2}{9}\)
\(\Leftrightarrow x=\frac{94}{99}\)
b)\(\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(2\left(\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{18}\)
\(x+1=18\)
\(x=17\)
1) a) A=\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{3}-\frac{1}{8}=\frac{5}{24}\)
c) C=\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(C=1-\frac{1}{101}\)
\(C=\frac{100}{101}\)
d) Sửa đề: thay \(\frac{3}{92.98}\)=\(\frac{3}{92.95}\)
\(D=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{92}-\frac{1}{95}\)
\(D=\frac{1}{2}-\frac{1}{95}\)
\(D=\frac{95-2}{190}=\frac{93}{190}\)
Các bài trên áp dụng theo tính chất: \(\frac{a}{b\left(b+a\right)}\frac{1}{b}-\frac{1}{b+a}\)
2x=1/11-1/13.........................-1/53-1/55+3/11
2x=1/11-1/55+3/11
2x=19/55
x=19/55 chia 2
x=19/110
Ta có: \(x-\frac{20}{11\cdot13}-\frac{20}{13\cdot15}-...-\frac{20}{53\cdot55}=\frac{3}{11}\)
\(\Leftrightarrow x-10\cdot\left(\frac{2}{11\cdot13}+\frac{2}{13\cdot15}+...+\frac{2}{53\cdot55}\right)=\frac{3}{11}\)
\(\Leftrightarrow x-10\cdot\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Leftrightarrow x-10\cdot\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(\Leftrightarrow x-10\cdot\frac{4}{55}=\frac{3}{11}\)
\(\Leftrightarrow x-\frac{8}{11}=\frac{3}{11}\)
\(\Leftrightarrow x=\frac{3}{11}+\frac{8}{11}\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)thỏa mãn đề.
