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Hello Triệu Mẫn điên .Tui là Nguyên 6n1^^
Tui đang suy nghĩ
Tui biết làm nhưng không nói
chỉ nói kết quả bằng 10
Ta có
\(2017-\left(\frac{1}{4}+\frac{2}{5}+\frac{3}{6}+\frac{4}{7}+...+\frac{2017}{2020}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{4}+\frac{2}{5}+...+\frac{2017}{2020}\right)\)
\(=\left(1-\frac{1}{4}\right)+\left(1-\frac{2}{5}\right)+...+\left(1-\frac{2017}{2020}\right)\)
\(=\frac{3}{4}+\frac{3}{5}+....+\frac{3}{2020}\)
\(=\frac{3.5}{4.5}+\frac{3.5}{5.5}+\frac{3.5}{6.5}+...+\frac{3.5}{2020.5}\)
\(=3.5\left(\frac{1}{4.5}+\frac{1}{5.5}+\frac{1}{6.5}+...+\frac{1}{2020.5}\right)\)
\(=15.\left(\frac{1}{20}+\frac{1}{25}+\frac{1}{30}+...+\frac{1}{10100}\right)\)
Thế vào ta có
\(\frac{15.\left(\frac{1}{20}+\frac{1}{25}+\frac{1}{30}+...+\frac{1}{10100}\right)}{\frac{1}{20}+\frac{1}{25}+...+\frac{1}{10100}}=15\)
Được cập nhật 41 giây trước (17:23)
Ta có :
2017−(14 +25 +36 +47 +...+20172020 )
=(1+1+...+1)−(14 +25 +...+20172020 )
=(1−14 )+(1−25 )+...+(1−20172020 )
=34 +35 +....+32020
=3.54.5 +3.55.5 +3.56.5 +...+3.52020.5
=3.5(14.5 +15.5 +16.5 +...+12020.5 )
=15.(1
\(\frac{19}{37}+\left(1-\frac{19}{37}\right)\)
\(=\frac{19}{37}+1-\frac{19}{37}\)
\(=\left(\frac{19}{37}-\frac{19}{37}\right)+1\)
\(=0+1=1\)
\(P=\frac{1}{99}-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{98.99}\right)\)
\(=\frac{1}{99}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(=\frac{1}{99}-\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{99}-\frac{98}{99}\)
\(=-\frac{97}{99}\)
Vậy \(P=-\frac{97}{99}\)
P=-1/1.2-1/2.3-...-1/98.99-1/99
P=-(1/1.2+1/2.3+...+1/98.99+1/99)
P=-1
Ta có: \(\frac{3}{1^2.2^2}=\frac{3}{1.4}=1-\frac{1}{4}\); \(\frac{5}{2^2.3^2}=\frac{5}{4.9}=\frac{1}{4}-\frac{1}{9}\); \(\frac{7}{3^2.4^2}=\frac{7}{9.16}=\frac{1}{9}-\frac{1}{16}\); ...; \(\frac{39}{19^2.20^2}=\frac{39}{361.400}=\frac{1}{361}-\frac{1}{400}\)
Gọi tổng đó là A => A=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{361}-\frac{1}{400}\)
=> \(A=1-\frac{1}{400}=\frac{399}{400}< \frac{400}{400}=1\)
=> A < 1
Đặt \(A=2017-\frac{1}{4}-\frac{2}{5}-...-\frac{2017}{2010}\)
\(B=\frac{1}{20}+\frac{1}{25}+\frac{1}{30}+...+\frac{1}{10100}\)
Ta có:
\(A=2017-\frac{1}{4}-\frac{2}{5}-...-\frac{2017}{2020}\)
\(A=1-\frac{1}{4}+1-\frac{2}{5}+1-\frac{3}{6}+...+1-\frac{2017}{2020}\)
\(A=\frac{3}{4}+\frac{3}{5}+\frac{3}{6}+...+\frac{3}{2020}\)
\(A=3\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2020}\right)\)
\(B=\frac{1}{20}+\frac{1}{25}+\frac{1}{30}+...+\frac{1}{10100}\)
\(B=\frac{1}{4.5}+\frac{1}{5.5}+\frac{1}{6.5}+...+\frac{1}{2020.5}\)
\(B=\frac{1}{5}\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2020}\right)\)
\(\frac{A}{B}=\frac{3\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2020}\right)}{\frac{1}{5}\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2020}\right)}=\frac{3}{\frac{1}{5}}=15\)
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https://olm.vn/hoi-dap/detail/217907126396.html