Tính nhanh
F = 1 phần 18 + 1 phần 54 + 1 phần 108 +...+ 1 phần 990
Help
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{75}{100}=\dfrac{3}{4};\dfrac{63}{126}=\dfrac{1}{2}\\ \dfrac{36}{27}=1\dfrac{1}{3};\dfrac{81}{54}=1\dfrac{1}{2};\dfrac{105}{405}=\dfrac{7}{27}\)
1,
\(\frac{25}{12}+\left(\frac{-4}{12}\right)=\frac{7}{4}\)
\(\frac{-10}{8}+\frac{15}{4}=\frac{5}{2}\)
\(\frac{3}{8}+\frac{-14}{6}=\frac{-47}{24}\)
\(\frac{350}{150}+\left(\frac{-200}{360}\right)=\frac{16}{9}\)
\([\frac{5}{8}+\left(\frac{-3}{4}\right)]+\frac{15}{6}=\frac{-1}{8}+\frac{15}{6}=\frac{19}{8}\)
\(\frac{7}{3}+[\left(\frac{-5}{6}\right)+\left(\frac{-2}{3}\right)]=\frac{7}{3}+\left(\frac{-3}{2}\right)=\frac{5}{6}\)
bài 4:so sánh
5/2 lớn hơn 3/7
4/3 lớn hơn,3/2 lớn hơn
bài 6:rút gọn các phân số sau:
3/9=1/3 9/12=3/4 8/18=4/9 60/36=10/6 17/34=1/2 17/51=1/3 35/100=7/20 25/100=1/4 8/1000=1/125 24/30=4/5 18/54=1/3 72/42=12/7
đay nhé mk chưa làm hết đc bn viết liền quá mk nhìn khó mà mk hỏi bài 7 là nhân hay cộng vậy?
\(\dfrac{4}{5}+\dfrac{19}{18}-\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{10}{8}\)
\(=\left(\dfrac{4}{5}+\dfrac{1}{5}\right)-\left(\dfrac{4}{8}+\dfrac{10}{8}\right)+\dfrac{19}{18}\)
\(=\dfrac{5}{5}-\dfrac{14}{8}+\dfrac{19}{18}\)
\(=1-\dfrac{7}{4}+\dfrac{19}{18}\)
\(=-\dfrac{3}{4}+\dfrac{19}{18}=\dfrac{11}{36}\)
\(\dfrac{4}{5}+\dfrac{19}{18}-\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{10}{8}=\dfrac{4}{5}+\dfrac{19}{18}-\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{5}{4}=\left(\dfrac{4}{5}+\dfrac{1}{5}\right)+\left(\dfrac{19}{18}-\dfrac{1}{2}\right)-\dfrac{5}{4}=1+\dfrac{5}{9}-\dfrac{5}{4}=\dfrac{36}{36}+\dfrac{20}{36}-\dfrac{45}{36}=\dfrac{11}{36}\)
a: \(=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{47}-\dfrac{4}{47}+\dfrac{9}{53}\right)}=\dfrac{3}{4}\)
b: \(F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(=2\cdot\dfrac{1004}{2010}=\dfrac{2008}{2010}=\dfrac{1004}{1005}\)
c: \(S=\dfrac{1}{3\cdot6}+\dfrac{1}{6\cdot9}+...+\dfrac{1}{30\cdot33}\)
\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{10}{33}=\dfrac{10}{99}\)
\(F=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(F=\frac{1}{9}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)\)
\(F=\frac{1}{9}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)
\(F=\frac{1}{9}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(F=\frac{1}{9}\left(1-\frac{1}{11}\right)\)
\(F=\frac{1}{9}.\frac{10}{11}=\frac{10}{99}\)