Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(S=\frac{3}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}\)
\(S=\frac{3}{1.2}-\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+\frac{11}{5.6}-\frac{13}{6.7}+\frac{15}{7.8}-\frac{17}{8.9}\)
\(S=3\left(\frac{1}{1.2}\right)-5\left(\frac{1}{2.3}\right)+7\left(\frac{1}{3.4}\right)-9\left(\frac{1}{4.5}\right)+11\left(\frac{1}{5.6}\right)-13\left(\frac{1}{6.7}\right)+15\left(\frac{1}{7.8}\right)-17\left(\frac{1}{8.9}\right)\)
\(S=3\left(1-\frac{1}{2}\right)-5\left(\frac{1}{2}-\frac{1}{3}\right)+7\left(\frac{1}{3}-\frac{1}{4}\right)-9\left(\frac{1}{4}-\frac{1}{5}\right)+11\left(\frac{1}{5}-\frac{1}{6}\right)-13\left(\frac{1}{6}-\frac{1}{7}\right)+15\left(\frac{1}{7}-\frac{1}{8}\right)-17\left(\frac{1}{8}-\frac{1}{9}\right)\)
\(S=\left(3-\frac{3}{2}\right)-\left(\frac{5}{2}-\frac{5}{3}\right)+\left(\frac{7}{3}-\frac{7}{4}\right)-\left(\frac{9}{4}-\frac{9}{5}\right)+\left(\frac{11}{5}-\frac{11}{6}\right)-\left(\frac{13}{6}-\frac{13}{7}\right)+\left(\frac{15}{7}-\frac{15}{8}\right)-\left(\frac{17}{8}-\frac{17}{9}\right)\)\(S=3-\frac{3}{2}-\frac{5}{2}+\frac{5}{3}+\frac{7}{3}-\frac{7}{4}-\frac{9}{4}+\frac{9}{5}+\frac{11}{5}-\frac{11}{6}-\frac{13}{6}+\frac{13}{7}+\frac{15}{7}-\frac{15}{8}-\frac{17}{8}+\frac{17}{9}\) Giờ bạn chỉ cần nhóm từng cặp phân số có cùng tử số rồi tính tiếp là ra kết quả thôi
( khi nhóm cặp nhớ đổi dấu nha)
\(A=\frac{3}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}\)
\(=\left(1+\frac{1}{2}\right)-\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)-\left(\frac{1}{4}+\frac{1}{5}\right)+...-\left(\frac{1}{8}+\frac{1}{9}\right)\)
\(=1+\frac{1}{2}-\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}-\frac{1}{4}-\frac{1}{5}+\frac{1}{5}+\frac{1}{6}-...-\frac{1}{9}\)
\(=1-\frac{1}{9}\)
\(=\frac{8}{9}\)
\(=\frac{3}{1.2}+\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+\frac{11}{5.6}-\frac{13}{6.7}+\frac{15}{7.8}-\frac{17}{8.9}\)
\(A=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{90}\right)\)
\(A=\left(1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)\)
\(A=9+\left(\frac{1}{1.2}+\frac{1}{2\cdot3}+\frac{1}{3.4}+...+\frac{1}{9\cdot10}\right)\)
\(A=9+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=9+\left(1-\frac{1}{10}\right)=9-\frac{9}{10}=8\frac{1}{10}\)
\(A=10.\left(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+....+\frac{71}{72}+\frac{89}{90}\right)\)
Đặt \(B=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{71}{72}+\frac{89}{90}\)
\(B=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+...+\left(1-\frac{1}{72}\right)+\left(1-\frac{1}{90}\right)\)
\(B=1+1+1+1+...+1-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{72}+\frac{1}{90}\right)\)
\(B=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(B=9-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(B=9-\left(\frac{1}{1}-\frac{1}{10}\right)=9-\frac{9}{10}=\frac{81}{10}=8,1\)
Ta có \(A=10.B=10.B=10.8,1=81\)
Vậy \(A=81\)
a/=(74-(-1937)1)
=74-(-1937)
=2011
b/=4/7+5/6:5-3/8*(-4)
=4/7+1/6-(-3/2)
=31/42-(-3/2)
=47/21
minh chi biet bay nhieu
S = \(\frac{3}{1.2}-\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+\frac{11}{5.6}-\frac{13}{6.7}+\frac{15}{7.8}-\frac{17}{8.9}\)
= \(\left(\frac{1}{1}+\frac{1}{2}\right)-\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)-...-\left(\frac{1}{8}+\frac{1}{9}\right)\) ( Vì \(\frac{a}{b.c}=\frac{1}{b}+\frac{1}{c}\)với b+c=a )
= \(\frac{1}{1}+\frac{1}{2}-\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}-\frac{1}{4}-...-\frac{1}{8}-\frac{1}{9}\)
= \(\frac{1}{1}-\frac{1}{9}\)
= \(\frac{8}{9}\)