Bài 1:Tính tổng
S=\(\frac{14}{3.5}+\frac{14}{5.7}+\frac{14}{7.9}+...+\frac{14}{61.63}\) so sánh với \(\frac{5}{2}\)
Help meeeeeeeeeeeeeeeee
Lớp 6
Tính tổng S = a +b cho biết : \(\frac{14}{1.3}+\frac{14}{3.5}+\frac{14}{5.7}+\frac{14}{7.11}\)\(+\frac{14}{11.13}+\frac{14}{13.15}\)
Nay rảnh quá đăng câu hỏi cho mn làm, gợi ý đáp số là S = 113
Sửa đề : v
\(S=\frac{14}{1.3}+\frac{14}{3.5}+\frac{14}{5.7}+\frac{14}{7.9}+...+\frac{14}{13.15}\)
\(S=7.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{13.15}\right)\)
\(S=7.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)
\(S=7.\left(1-\frac{1}{15}\right)\)
\(S=7.\frac{14}{15}=\frac{98}{15}\)
mình tính cái đa thức ở sau nhé:
=14.(1/1.3+1/3.5+...+1/13.15)
=7.(1-1/3+1/3-1/5+...+1/13-1/15)
=7.(1-1/15)
=7.(14/15)
=98/15
còn a,b là gì thì mình ko bt
HGH̃TUYGHUGŨYHJGGBGJ
Tính hợp lí:
\(a)\frac{0,4-\frac{2}{9}+\frac{2}{11}}{0,6-\frac{3}{9}+\frac{3}{11}}+\frac{\frac{2}{3}+\frac{2}{7}-\frac{1}{14}}{-1-\frac{3}{7}+\frac{3}{28}}\)
\(b)1-\frac{2}{3.5}-\frac{2}{5.7}-\frac{2}{7.9}-.....-\frac{2}{97.99}\)
Giúp mình với! Mình cần gấp!
Ai nhanh mình tích cho!
a, Ta có:
\(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{0,6-\frac{3}{9}+\frac{3}{11}}+\frac{\frac{2}{3}+\frac{2}{7}-\frac{1}{14}}{-1-\frac{3}{7}+\frac{3}{28}}=\frac{2\left(0,2-\frac{1}{9}+\frac{1}{11}\right)}{3\left(0,2-\frac{1}{9}+\frac{1}{11}\right)}+\frac{2\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{28}\right)}{-3\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{28}\right)}=\frac{2}{3}+\frac{-2}{3}=0\)
k đúng cho mình nha. Thanks!!!
a, bày cho mình cách viết bằng phân số đi , mình trình bày cách làm cho. k đúng cho mình nha.
\(=\frac{2}{3}\times\frac{0,1-\frac{1}{9}+\frac{1}{11}}{0,1-\frac{1}{9}+\frac{1}{11}}+\frac{-\frac{2}{3}\times\left(-1-\frac{3}{7}+\frac{3}{28}\right)}{-1-\frac{3}{7}+\frac{3}{28}}\)
=\(\frac{2}{3}+\left(-\frac{3}{2}\right)\)
=\(-\frac{5}{6}\)
tìm x biết: \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{19.21}-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{19\cdot21}-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{19\cdot21}\right)-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{1}{2}\left(1-\frac{1}{21}\right)-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{1}{2}\cdot\frac{20}{21}-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{10}{21}-\frac{x}{14}=\frac{2}{-7}\)
\(\frac{x}{14}=\frac{10}{21}-\frac{2}{-7}\)
\(\frac{x}{14}=\frac{16}{21}\)
\(\Rightarrow x\cdot=21=14\cdot16\)
\(\Rightarrow x\cdot21=224\)
\(\Rightarrow x=\frac{224}{21}\)
a) \(A=\frac{1}{199}-\frac{1}{199.198}-\frac{1}{198.197}-\frac{1}{197.196}-....-\frac{1}{3.2}-\frac{1}{2.1}\)
b) \(B=1-\frac{2}{3.5}-\frac{2}{5.7}-\frac{2}{7.9}-....-\frac{2}{61.63}-\frac{2}{63-65}\)
\(a)\) \(A=\frac{1}{199}-\frac{1}{199.198}-\frac{1}{198.197}-\frac{1}{197.196}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(A=\frac{1}{199}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{197.198}+\frac{1}{198.199}\right)\)
\(A=\frac{1}{199}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{197}-\frac{1}{198}+\frac{1}{198}-\frac{1}{199}\right)\)
\(A=\frac{1}{199}-\left(1-\frac{1}{199}\right)\)
\(A=\frac{1}{199}-1+\frac{1}{199}\)
\(A=\frac{-197}{199}\)
Chúc bạn học tốt ~
a) \(A=\frac{1}{199}-\frac{1}{199.198}-\frac{1}{198.197}-\frac{1}{197.196}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(A=\frac{1}{199}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...\frac{1}{197.198}+\frac{1}{198.199}\right)\)
\(A=\frac{1}{199}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{197}-\frac{1}{198}+\frac{1}{198}-\frac{1}{199}\right)\)
\(A=\frac{1}{199}-(1-\frac{1}{199})\)
\(A=\frac{1}{199}-1+\frac{1}{199}\)
\(A=\frac{-197}{199}\)
Vậy \(A=\frac{-197}{199}\)
b) cũng thế
Tính hợp lí:
a, 75. ( \(-2\frac{3}{25}+7\frac{2}{75}-5\frac{4}{15}\) )
b, \(45.\left(5\frac{4}{15}-4\frac{7}{9}-1\frac{8}{45}\right)\)
c, \(\frac{-5}{8}+\frac{14}{18}-\frac{3}{8}+\frac{2}{9}-\frac{1}{2006}\)
d, \(\frac{15}{29}-\frac{8}{7}+\frac{16}{14}+\frac{14}{29}-\frac{3}{8}\)
e, \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
Help: Cho A=\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\).Hãy so sánh A với \(\frac{1}{2}\)
Ta có:
\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)
\(=\frac{1}{4}+\left(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\right)\)
Đặt \(B=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)
\(B=\left(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}\right)+\left(\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\right)\)
Giả sử tất cả các số hạng của B đều bằng \(\frac{1}{6^2}\)
\(\Rightarrow B=6.\frac{1}{6^2}=\frac{6}{36}=\frac{1}{6}<\frac{1}{4}\)
Do đó \(B<\frac{1}{4}\)
\(\Rightarrow A=\frac{1}{4}+B<\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\)
Vậy \(A<\frac{1}{2}\)
So sánh bằng cách nhanh nhất(ko quy đồng)
a.A=\(\frac{13+14}{14+5}\)và \(\frac{13}{14}+\frac{14}{15}\)
b.So sánh B với 2:B=\(\frac{2011}{2012}+\frac{2011}{2010}\)
c. So sánh K với\(\frac{3}{8}\):K=\(\frac{3x5+6x10+9x15+12x20+15x25}{5x7+10x14+15x21+20x28+25x35}\)
tính hợp lí
\(\frac{11}{125}-\frac{17}{18}-\frac{5}{7}+\frac{4}{9}+\frac{17}{14}\)
\(\frac{42}{46}+\frac{250}{286}+\frac{-2121}{2323}+\frac{-125125}{143143}\)
tính
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2003.2004}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{2003.2005}\)
Bài 1: Thực hiện phép tính
\(\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{0,75+\frac{9}{7}-2\frac{2}{5}}+\frac{\frac{3}{14}-\frac{2}{10}+\frac{5}{18}+\frac{7}{66}}{\frac{6}{7}-\frac{4}{5}+\frac{10}{9}+\frac{14}{13}}\)
Vừa thi về, giải đc ùi nhưng muốn xem k quả của các bạn
Mình làm như thế này nek
\(\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{0,75+\frac{9}{7}-2\frac{2}{5}}+\frac{\frac{3}{14}-\frac{2}{10}+\frac{5}{18}+\frac{7}{66}}{\frac{6}{7}-\frac{4}{5}+\frac{10}{9}+\frac{14}{33}}\)
\(=\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{\frac{2}{4}+\frac{9}{7}-\frac{12}{5}}+\frac{\frac{1}{2}\cdot\left(\frac{3}{7}-\frac{2}{5}+\frac{5}{9}+\frac{7}{33}\right)}{2\cdot\left(\frac{3}{7}-\frac{2}{5}+\frac{5}{9}+\frac{7}{33}\right)}\)
\(=\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{3\cdot\left(\frac{1}{4}+\frac{3}{7}-\frac{4}{5}\right)}+\frac{\frac{1}{2}}{2}\)
\(=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)