Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a.\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\)
\(=\frac{1}{2}-\frac{1}{5}\)
\(=\frac{3}{10}\)
\(b.\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}\)
\(=2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}\right)\)
\(=2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\right)\)
\(=2\cdot\left(\frac{1}{2}-\frac{1}{5}\right)\)
\(=2\cdot\frac{3}{10}=\frac{3}{5}\)
\(c.\frac{1}{2\cdot3}+\frac{2}{3\cdot5}+\frac{3}{5\cdot8}\)
\(=\frac{1}{6}+\frac{2}{15}+\frac{3}{40}\)
\(=\frac{3}{8}\)
k nha 500 AE
a, \(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}\)
\(=\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+\frac{5-4}{4\times5}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\)
\(=\frac{1}{2}-\frac{1}{5}\)
\(=\frac{3}{10}\)
b, \(\frac{2}{2\times3}+\frac{2}{3\times4}+\frac{2}{4\times5}\)
\(=\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+\frac{5-4}{4\times5}\)
\(=\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\right)\times\frac{2}{1}\)
\(=\left(\frac{1}{2}-\frac{1}{5}\right)\times\frac{2}{1}\)
\(=\frac{3}{10}\times\frac{2}{1}\)
\(=\frac{3}{5}\)
c, \(\frac{1}{2\times3}+\frac{2}{3\times5}+\frac{3}{5\times8}\)
\(=\frac{3-2}{2\times3}+\frac{5-3}{3\times5}+\frac{8-5}{5\times8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}\)
\(=\frac{3}{8}\)
ta có : A=1/2+1/4+..+1/1024
=> A=1/21+1/22+..+1/210
=> A.2=(1/21+1/22+..+1/210).2
=> A.2=1+1/21+1/22+..+1/29
=> 2A-A=(1+1/21+1/22+..+1/29)-(1/21+1/22+..+1/210)
=> A=1-1/210
\(a,\frac{4}{5}\times\frac{a}{b}=\frac{7}{10}\)
\(\frac{a}{b}=\frac{7}{10}:\frac{4}{5}\)
\(\frac{a}{b}=\frac{7}{10}\times\frac{5}{4}\)
\(\frac{a}{b}=\frac{7}{8}\)
\(b,\frac{1}{3}:\frac{a}{b}=\frac{2}{3}:\frac{4}{3}\)
\(\frac{1}{3}:\frac{a}{b}=\frac{2}{3}\times\frac{3}{4}\)
\(\frac{1}{3}:\frac{a}{b}=\frac{1}{2}\)
\(\frac{a}{b}=\frac{1}{3}:\frac{1}{2}\)
\(\frac{a}{b}=\frac{1}{3}\times2\)
\(\frac{a}{b}=2\)
\(c,\frac{a}{b}+\frac{3}{8}=\frac{5}{6}\)
\(\frac{a}{b}=\frac{5}{6}-\frac{3}{8}\)
\(\frac{a}{b}=\frac{11}{24}\)
bài 1
Ta có : 2016/2017<1
2017/2018<1
Nên 2016/2017=2017/2018
Bài 1 :
a) Ta có : \(\frac{2016}{2017}=1-\frac{1}{2017}\)
\(\frac{2017}{2018}=1-\frac{1}{2018}\)
Vì \(-\frac{1}{2017}< -\frac{1}{2018}\)nên \(\frac{2016}{2017}< \frac{2017}{2018}\)
b) Ta có : \(\frac{2018}{2017}=1+\frac{1}{2017}\)
\(\frac{2017}{2016}=1+\frac{1}{2016}\)
Vì \(\frac{1}{2017}< \frac{1}{2016}\) nên \(\frac{2018}{2017}< \frac{2017}{2016}\)
Câu 2 :
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{101.103}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{101.103}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{103}\right)\)
\(=\frac{1}{2}.\frac{102}{103}=\frac{51}{103}\)
tính nhanh :
C= \(\frac{1}{3}\)+\(\frac{1}{10}\)+\(\frac{1}{15}\)+..................+\(\frac{1}{45}\)
C=\(\frac{1}{3}+\frac{1}{10}+\frac{1}{15}+\frac{1}{20}+\frac{1}{25}+\frac{1}{30}+\frac{1}{35}+\frac{1}{40}\)
=\(\frac{1}{3}+\frac{1}{15}+\frac{1}{10}+\frac{1}{20}+\frac{1}{30}+\frac{1}{40}+\frac{1}{25}+\frac{1}{35}\)
=\(\frac{5}{15}+\frac{1}{15}+\frac{4}{40}+\frac{2}{40}+\frac{1}{40}+\frac{1}{30}+\frac{1}{25}+\frac{1}{35}\)
=\(\frac{6}{15}+\frac{7}{40}+\frac{107}{1050}\)
a)
\(\frac{3\times4\times7}{5\times3\times4}=\frac{7}{5}\)
b)
\(\frac{2\times5\times6\times8}{6\times2\times8\times9}=\frac{5}{9}\)
c)
\(\frac{4\times5\times6}{10\times3\times8}=\frac{4\times5\times3\times2}{5\times2\times3\times4\times2}=\frac{1}{2}\)
\(a,\frac{3\times4\times7}{5\times3\times4}=\frac{7}{5}\)
\(b,\frac{2\times5\times6\times8}{6\times2\times8\times9}=\frac{5}{9}\)