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Bài 1 mik học xong quên hết òi (mấy bài kia là hok biết luôn :V)
Đặt \(B=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{210}\)
\(\frac{1}{2}B=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}\)
\(\frac{1}{2}B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{20.21}\)
\(\frac{1}{2}B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{20}-\frac{1}{21}\)
\(\frac{1}{2}B=\frac{1}{2}-\frac{1}{21}\)
\(\Rightarrow B=\frac{\frac{1}{2}-\frac{1}{21}}{\frac{1}{2}}=\frac{19}{21}\)
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+3+...+50}\)
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{\frac{\left(1+50\right).50}{2}}\)
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{1275}\)
\(A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{2550}\)
\(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+..+\frac{2}{50.51}\)
\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{51}\right)=2\cdot\frac{49}{102}=\frac{49}{51}\)
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)\(\frac{2}{42}+\frac{2}{56}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(2\left(\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{18}\)
\(x+1=18\)
\(x=17\)
có : Q = [ 2 + 2^2 ] + [ 2^3 +2^4] + ... + [2^9 + 2^10]
Q = 2 [1+2] +2^3[1 +2]+ ...+ 2^9 [1+2]
Q = 2 . 3+2^3 .3 +... + 2^9 .3
Q = 3. [ 2 + 2^3 +... + 2^9]
Vậy Q chia hết cho 3
a, \(A=\frac{2}{5}+\frac{-1}{6}-\frac{3}{4}-\frac{-2}{3}\)
\(A=\left(\frac{2}{5}-\frac{3}{4}\right)+\left(\frac{-1}{6}-\frac{-2}{3}\right)\)
\(A=\left(\frac{8}{20}-\frac{15}{20}\right)+\left(\frac{-3}{18}-\frac{-12}{18}\right)\)
\(A=\frac{-7}{20}+\frac{1}{2}\)
\(\Rightarrow A=\frac{-7}{20}+\frac{10}{20}=\frac{3}{20}\)
b, \(B=\frac{7}{10}-\frac{-3}{4}+\frac{-5}{6}-\frac{1}{5}+\frac{-2}{3}\)
\(B=\left(\frac{7}{10}-\frac{1}{5}\right)+\left(\frac{-5}{6}+\frac{-2}{3}\right)-\frac{-3}{4}\)
\(B=\left(\frac{7}{10}-\frac{2}{10}\right)+\left(\frac{-5}{6}+\frac{-4}{6}\right)-\frac{-3}{4}\)
\(B=\frac{1}{2}+\frac{-3}{2}-\frac{-3}{4}\)
\(B=\frac{2}{4}+\frac{-6}{4}-\frac{-3}{4}\)
\(\Rightarrow B=\frac{2+-6+3}{4}=\frac{-1}{4}\)
c, \(C=\frac{\left(\frac{1}{2}-0,75\right)\times\left(0,2-\frac{2}{5}\right)}{\frac{5}{9}-1\frac{1}{12}}\)
\(C=\frac{\left(\frac{1}{2}-\frac{3}{4}\right)\times\left(\frac{1}{5}-\frac{2}{5}\right)}{\frac{5}{9}-\frac{1\times12+1}{12}}\)
\(C=\frac{\left(\frac{2}{4}-\frac{3}{4}\right)\times\left(\frac{-1}{5}\right)}{\frac{5}{9}-\frac{13}{12}}\)
\(C=\frac{\left(\frac{-1}{4}\right)\times\left(\frac{-1}{5}\right)}{\frac{60}{108}-\frac{117}{108}}\)
\(C=\frac{\frac{1}{20}}{\frac{-19}{36}}=\frac{1}{20}\div\frac{-19}{36}=\frac{1}{20}\times\frac{36}{-19}\)
\(\Rightarrow C=\frac{36}{-380}=\frac{-9}{95}\)
d, \(D=\frac{\frac{2}{3}+\frac{2}{7}-\frac{1}{4}}{-1-\frac{3}{7}+\frac{3}{28}}\)
\(D=\frac{\frac{56}{84}+\frac{24}{84}-\frac{21}{84}}{\frac{-10}{7}+\frac{3}{28}}\)
\(D=\frac{\frac{59}{84}}{\frac{-40}{28}+\frac{2}{28}}=\frac{59}{84}\div\frac{-37}{28}=\frac{59}{84}\times\frac{28}{-37}\)
\(\Rightarrow D=\frac{1652}{-3108}=\frac{-59}{111}\)
Bài nhìn vô muốn xỉu rồi ='((
1. a) \(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{91.94}+\frac{2}{94.97}\)
\(=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{91.94}+\frac{3}{94.97}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{97}\right)=\frac{2}{3}.\frac{96}{97}=\frac{64}{97}\)
b) Bạn tự làm, làm nữa chắc xỉu =((( Khi nào rảnh mình sẽ làm, nếu bạn cần
2 )
a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{1005}{2011}\)
\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)
\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{1005}{2011}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{1005}{2011}:2=\frac{1005}{4022}\)
\(\Leftrightarrow\frac{1}{x+2}=1-\frac{1005}{4022}=\frac{3017}{4020+2}\)
\(\Rightarrow x=4020\)
\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\)
\(\frac{1}{2}A=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{11}}\)
\(A=\left(a-\frac{1}{2}A\right).2=\left(\frac{1}{2}-\frac{1}{2^{11}}\right).2=\frac{1023}{1024}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{10}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^9}\)
\(2A-A=1-\frac{1}{2^{10}}\)