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Bài 1: Tính tổng 100 số hạng đầu tiên của các dãy sau:
a) \(\left\{{}\begin{matrix}\dfrac{1}{2}=\dfrac{1}{1.2}\\\dfrac{1}{6}=\dfrac{1}{2.3}\\\dfrac{1}{12}=\dfrac{1}{3.4}\\...\end{matrix}\right.\)
Vậy số thứ 100 của dãy là: \(\dfrac{1}{100.101}=\dfrac{1}{10100}\)
Tổng: \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{100.101}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{100}-\dfrac{1}{101}\)
\(=1-\dfrac{1}{101}\)
\(=\dfrac{100}{101}\)
b) \(\left\{{}\begin{matrix}\dfrac{1}{6}=\dfrac{1}{\left(5.0+1\right)\left(5.1+1\right)}\\\dfrac{1}{66}=\dfrac{1}{\left(5.1+1\right)\left(5.2+1\right)}\\\dfrac{1}{176}=\dfrac{1}{\left(5.2+1\right)\left(5.3+1\right)}\\...\end{matrix}\right.\)
Vậy số thứ 100 của dãy là: \(\dfrac{1}{\left(5.99+1\right)\left(5.100+1\right)}=\dfrac{1}{248496}\)
Tổng: \(\dfrac{1}{1.6}+\dfrac{1}{6.11}+\dfrac{1}{11.16}+...+\dfrac{1}{496.501}\)
\(=\dfrac{1}{5}\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+...+\dfrac{5}{496.501}\right)\)
\(=\dfrac{1}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{496}-\dfrac{1}{501}\right)\)
\(=\dfrac{1}{5}\left(1-\dfrac{1}{501}\right)\)
\(=\dfrac{1}{5}.\dfrac{500}{501}\)
\(=\dfrac{100}{501}\)
Bài 2: Tính:
a) \(A=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{97.3}+\dfrac{1}{99.1}}\)
\(A=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{3}+\dfrac{1}{97}\right)+...+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}\)
\(A=\dfrac{\dfrac{100}{1.99}+\dfrac{100}{3.97}+\dfrac{100}{5.95}+...+\dfrac{100}{49.51}}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}\)
\(A=\dfrac{100\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{49.51}\right)}\)
\(\Rightarrow A=\dfrac{100}{2}=50\)
Mình nghĩ đề là : 2/8 sẽ hay hơn.
\(B=\dfrac{5}{2}+\dfrac{6}{11}+\dfrac{2}{8}+\dfrac{7}{2}+\dfrac{6}{8}+\dfrac{5}{11}\)
\(=\left(\dfrac{5}{2}+\dfrac{7}{2}\right)+\left(\dfrac{6}{11}+\dfrac{5}{11}\right)+\left(\dfrac{2}{8}+\dfrac{6}{8}\right)\)
\(=6+1+1=8\)
= \(\dfrac{5}{2}(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2019}-\dfrac{1}{2021})\)
= \(\dfrac{5}{2}\left(1-\dfrac{1}{101}\right)\)
= \(\dfrac{5}{2}.\dfrac{100}{101}\)
= \(\dfrac{250}{101}\)
\(=\dfrac{5\left(5+2\right)}{11.7}+\dfrac{6}{11}=\dfrac{5}{11}+\dfrac{6}{11}=1\)
Ta có:
= 5/11 .( 5/7+ 2/7 ) + 6/11
= 5/11 . 1 + 6/11
= 11/ 11
= 1
#học tốt ạ :3
a: \(=\dfrac{-2}{12}-\dfrac{5}{12}+\dfrac{7}{12}=0\)
b: \(=\left(\dfrac{4}{45}-\dfrac{1}{45}+\dfrac{7}{45}+\dfrac{4}{45}-\dfrac{2}{45}-\dfrac{9}{45}\right)=\dfrac{3}{45}=\dfrac{1}{15}\)
\(\dfrac{6}{13}.\dfrac{5}{7}+\dfrac{6}{13}.\dfrac{2}{7}+\dfrac{7}{13}=\dfrac{6}{13}.\left(\dfrac{5}{7}+\dfrac{2}{7}\right)+\dfrac{7}{13}=\dfrac{6}{13}.1+\dfrac{7}{13}=\dfrac{6}{13}+\dfrac{7}{13}=\dfrac{13}{13}=1\)
#Monster
\(\dfrac{6}{13}\).(\(\dfrac{2}{7}+\dfrac{5}{7}\))+\(\dfrac{7}{13}\)
\(\dfrac{6}{13}\).1+\(\dfrac{7}{13}\)
\(\dfrac{6}{13}\)+\(\dfrac{7}{13}\)
1
\(\dfrac{-1}{9}.\dfrac{-3}{5}+\dfrac{5}{-6}.\dfrac{-3}{5}-\dfrac{7}{2}.\dfrac{3}{5}\)
\(=\dfrac{3}{5}.\left(\dfrac{1}{9}+\dfrac{5}{6}-\dfrac{7}{2}\right)\)
\(=\dfrac{3}{5}.\left(\dfrac{2}{18}+\dfrac{15}{18}-\dfrac{63}{18}\right)\)
\(=\dfrac{3}{5}.\left(-\dfrac{23}{9}\right)\)
\(=-\dfrac{69}{45}\)
\(=\left(\dfrac{2}{7}+\dfrac{5}{7}\right)+\left(\dfrac{3}{11}-\dfrac{6}{11}\right)+\dfrac{1}{4}=\dfrac{5}{4}-\dfrac{3}{11}=\dfrac{55}{44}-\dfrac{12}{44}=\dfrac{43}{44}\)
`2/7-(-3)/11+5/7-1/(-4)-6/11`
`=2/7+3/11+5/7+1/4-6/11`
`=(2/7+5/7)+(3/11-6/11)+1/4`
`=7/7-3/11+1/4`
`=1-3/11+1/4`
`=44/44-12/44+11/44`
`=43/44`
Giải:
\(\dfrac{5}{6}+\dfrac{5}{66}+\dfrac{5}{176}+\dfrac{5}{336}\)
\(=5\left(\dfrac{1}{6}+\dfrac{1}{66}+\dfrac{1}{176}+\dfrac{1}{336}\right)\)
\(=5.\dfrac{4}{21}\)
\(=\dfrac{20}{21}\)
Vậy ...