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![](https://rs.olm.vn/images/avt/0.png?1311)
Hello Triệu Mẫn điên .Tui là Nguyên 6n1^^
Tui đang suy nghĩ
Tui biết làm nhưng không nói
chỉ nói kết quả bằng 10
![](https://rs.olm.vn/images/avt/0.png?1311)
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:
\(A=2017-\frac{1}{4}-\frac{2}{5}-...-\frac{2017}{2020}\)
\(A=1-\frac{1}{4}+1-\frac{2}{5}+1-\frac{3}{6}+...+1-\frac{2017}{2020}\)
\(A=\frac{3}{4}+\frac{3}{5}+\frac{3}{6}+...+\frac{3}{2020}\)
\(A=3\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2020}\right)\)
\(B=\frac{1}{20}+\frac{1}{25}+\frac{1}{30}+...+\frac{1}{10100}\)
\(B=\frac{1}{4.5}+\frac{1}{5.5}+\frac{1}{6.5}+...+\frac{1}{2020.5}\)
\(B=\frac{1}{5}\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2020}\right)\)
\(\frac{A}{B}=\frac{3\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2020}\right)}{\frac{1}{5}\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2020}\right)}=\frac{3}{\frac{1}{5}}=15\)
![](https://rs.olm.vn/images/avt/0.png?1311)
C\(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{5.6}\)-\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)-\(\frac{1}{8.9}+\frac{1}{9.10}\)
c=\(\frac{1}{1}-\frac{1}{10}\)
c=\(\frac{9}{10}\)
còn a và b rễ lắm mình ko thích làm bài rễ đâu bạn cố chờ lời giải khác nhé!
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\frac{12}{7}\cdot\frac{3}{4}-\frac{6}{7}\cdot\frac{4}{3}+\frac{6}{7}\)
\(=\frac{6}{7}\left(\frac{3}{2}-\frac{4}{3}+1\right)\)
\(=\frac{6}{7}\left(\frac{1}{6}+1\right)=\frac{6}{7}\cdot\frac{7}{6}=1\)
2.
\(=2017\cdot2018\cdot\left[\left(2016\cdot2018\right)-\left(2016\cdot2017\right)\right]\)
\(=2017\cdot2018\cdot2016\left(2018-2017\right)=2016\cdot2017\cdot2018\)
3.
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{100}-1\right)=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{99}{100}\)
\(=\frac{1}{100}\)
4.
\(=\frac{1+2+2^2+2^4+...+2^9}{2\left(1+2+2^2+2^3+2^4+...+2^9\right)}\)
\(=\frac{1}{2}\)
mình chỉ làm được câu 3 thôi
có \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)....\left(\frac{1}{100}-1\right)\)
\(=\frac{-1}{2}\times\frac{-2}{3}\times....\times\frac{-99}{100}\)
\(=\frac{\left(-1\right)\left(-2\right)....\left(-99\right)}{2\times3\times....\times100}\)
\(=\frac{-\left(1\times2\times....\times99\right)}{2\times3\times....\times100}\)
\(=\frac{-1}{100}\)
Ta có: 2017 -1/4 -2/5 -3/6 -... -2017/2020
= (1-1/4)+(1-2/5)+(1-3/6)+...+(1-2017/2020)
= 3/4 + 3/5 + 3/6 +...+ 3/2020
= 15 (1/20+ 1/25+ 1/30+...+ 1/10100)
Vậy B = 15.
Chúc bạn học tốt.