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Gọi dãy trên là A
\(\Leftrightarrow2A=\frac{2}{11\cdot13}+\frac{2}{13\cdot15}+...+\frac{2}{19\cdot21}\)
\(\Leftrightarrow2A=\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{19}-\frac{1}{21}\)
\(\Leftrightarrow2A=\frac{1}{11}-\frac{1}{21}+0+...+0\)
\(\Leftrightarrow2A=\frac{10}{231}\)
\(\Leftrightarrow A=\frac{5}{231}\)
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.
a, \(\frac{2020.125+1000}{126.2020-1020}=\frac{2020.125+1000}{2020.125+2020-1020}=1\)
b,\(\left(\frac{1}{11.13}+\frac{1}{13.15}+...+\frac{1}{19.21}\right).426+x=19\)
\(< =>\left(\frac{1}{11}-\frac{1}{21}\right).213+x=19\)
\(< =>\frac{2130}{231}+x=19\)
\(< =>x=19-\frac{2130}{231}=...\)
a)\(\frac{2020\cdot125+1000}{126\cdot2020-1020}=\frac{2020\cdot125+1000}{2020\cdot125+2020-1020}=\frac{2020\cdot125+1000}{2020\cdot125+1000}=1\)
b) \(\left(\frac{1}{11\cdot13}+\frac{1}{13\cdot15}+\frac{1}{15\cdot17}+\frac{1}{17\cdot19}+\frac{1}{19\cdot21}\right)\cdot426+x=19\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{2}{11\cdot13}+\frac{2}{13\cdot15}+\frac{2}{15\cdot17}+\frac{2}{17\cdot19}+\frac{2}{19\cdot21}\right)\cdot426+x=19\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+\frac{1}{17}-\frac{1}{19}+\frac{1}{19}-\frac{1}{21}\right)\cdot426+x=19\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{11}-\frac{1}{21}\right)\cdot426+x=19\)
\(\Leftrightarrow\frac{1}{2}\cdot\frac{10}{231}\cdot426+x=19\)
\(\Leftrightarrow\frac{710}{77}+x=19\)
\(\Leftrightarrow x=19-\frac{710}{77}=\frac{753}{77}\)
Sao đang phép trừ thành phép cộng vậy bạn. Nếu cọng hết thì mik bik tính đó.
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)\(=1008\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(=1008\cdot\left(1-\frac{1}{2017}\right)\)
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)\(\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(2\left(\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{18}\)
\(x+1=18\)
\(x=17\)
\(M=\frac{3}{2}.\left(\frac{2}{11.13}+\frac{2}{13.15}+......+\frac{2}{97.99}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+.....+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{11}-\frac{1}{99}\right)=\frac{3}{2}.\frac{8}{99}=\frac{4}{33}\)
M= \(\frac{3}{11\cdot13}+\frac{3}{13\cdot15}+\frac{3}{15\cdot17}+...+\frac{3}{97\cdot99}\)
=\(\frac{3}{2}\cdot\left(\frac{2}{11\cdot13}+\frac{2}{13\cdot15}+\frac{2}{15\cdot17}+...+\frac{2}{97\cdot99}\right)\)
=\(\frac{3}{2}\cdot\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{97}-\frac{1}{99}\right)\)
=\(\frac{3}{2}\cdot\left(\frac{1}{11}-\frac{1}{99}\right)\)
=\(\frac{3}{2}\cdot\frac{8}{99}\)
= \(\frac{4}{33}\)