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\(a.\frac{3}{7}.\frac{5}{6}+\frac{5}{6}.\frac{4}{7}=\left(\frac{3}{7}+\frac{4}{7}\right).\frac{5}{6}=1.\frac{5}{6}=\frac{5}{6}\)
\(b.\frac{1}{6}:\frac{4}{5}+\frac{1}{8}:\frac{4}{5}=\left(\frac{1}{6}+\frac{1}{8}\right):\frac{4}{5}=\left(\frac{4}{24}+\frac{3}{24}\right):\frac{4}{5}=\frac{7}{24}:\frac{4}{5}=\frac{7.5}{24.4}\)\(=\frac{35}{96}\)
\(c.\frac{1}{6}:\frac{4}{5}-\frac{1}{8}:\frac{4}{5}=\left(\frac{1}{6}-\frac{1}{8}\right):\frac{4}{5}=\left(\frac{4}{24}-\frac{3}{24}\right):\frac{4}{5}\)\(=\frac{1}{24}:\frac{4}{5}=\frac{1.5}{24.4}=\frac{5}{96}\)
a) \(\frac{3}{7}\cdot\frac{5}{6}+\frac{5}{6}\cdot\frac{4}{7}\)
\(=\left(\frac{3}{7}+\frac{4}{7}\right)\cdot\frac{5}{6}\)
\(=1\cdot\frac{5}{6}\)
\(=\frac{5}{6}\)
b) \(\frac{1}{6}:\frac{4}{5}+\frac{1}{8}:\frac{4}{5}\)
\(=\left(\frac{1}{6}+\frac{1}{8}\right):\frac{4}{5}\)
\(=\frac{7}{24}:\frac{4}{5}\)
\(=\frac{35}{96}\)
c) \(\frac{1}{6}:\frac{4}{5}-\frac{1}{8}:\frac{4}{5}\)
\(=\left(\frac{1}{6}-\frac{1}{8}\right):\frac{4}{5}\)
\(=\frac{1}{24}:\frac{4}{5}\)
\(=\frac{5}{96}\)
(\(\frac{1}{2}\):\(\frac{1}{3}\):\(\frac{1}{6}\)) +\(\frac{1}{5}\)
= 9 + \(\frac{1}{5}\)
=\(\frac{46}{5}\)
K minh nhe ban
\(\frac{1}{2}:\frac{1}{5}+\frac{1}{3}:\frac{1}{5}+\frac{1}{6}:\frac{1}{5}\)
\(=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right):\frac{1}{5}\)
\(=\left(\frac{3}{6}+\frac{2}{6}+\frac{1}{6}\right):\frac{1}{5}\)
\(=1:\frac{1}{5}\)
\(=1\times5\)
\(=5\)
Tính nhanh:
\(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\)\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}\)
\(=\left(\frac{1}{1}+\frac{1}{9}\right)+\left(\frac{1}{2}+\frac{1}{8}\right)\)\(+\left(\frac{1}{3}+\frac{1}{7}\right)+\left(\frac{1}{4}+\frac{1}{6}\right)+\frac{1}{5}\)
\(=\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{5}\)
\(=\frac{4}{10}+\frac{2}{5}=\frac{2}{5}+\frac{1}{5}=\frac{3}{5}\)
tks giúp mk nha! cảm ơn nhiều ạ...
Đặt \(A=2-1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(=2-\frac{1}{9}=\frac{18}{9}-\frac{1}{9}=\frac{17}{9}\)
\(\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{5}\right)\times\left(1-\frac{1}{6}\right)\times\left(1-\frac{1}{7}\right)\times\left(1-\frac{1}{8}\right)-\frac{1}{4}\times\frac{1}{2}\)
\(=\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times\frac{5}{6}\times\frac{6}{7}\times\frac{7}{8}-\frac{1}{4}\times\frac{1}{2}\)
\(=\frac{2}{8}-\frac{1}{4}\times\frac{1}{2}\)
\(=\frac{2}{8}-\frac{1}{8}=\frac{1}{8}\)
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 :
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{101.103}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{101.103}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{103}\right)\)
\(=\frac{1}{2}.\frac{102}{103}=\frac{51}{103}\)
\(\frac{3}{7}+\frac{4}{5}+\frac{4}{7}=\left(\frac{3}{7}+\frac{4}{7}\right)+\frac{4}{5}=1+\frac{4}{5}=\frac{9}{5}\)
\(\frac{6}{11}+\frac{1}{3}+\frac{5}{11}=\left(\frac{6}{11}+\frac{5}{11}\right)+\frac{1}{3}=1+\frac{1}{3}=\frac{4}{3}\)
a) \(\frac{3}{7}+\frac{4}{5}+\frac{4}{7}\)
= \(\left(\frac{3}{7}+\frac{4}{7}\right)+\frac{4}{5}\)
= \(1+\frac{4}{5}\)
= \(\frac{9}{5}\)
b) \(\frac{6}{11}+\frac{1}{3}+\frac{5}{11}\)
= \(\left(\frac{6}{11}+\frac{5}{11}\right)+\frac{1}{3}\)
= \(1+\frac{1}{3}\)
= \(\frac{4}{3}\)
1 \(A=\left(1+\frac{1}{2}\right)\times\left(1+\frac{1}{3}\right)\times\left(1+\frac{1}{4}\right)\times.........\times\left(1+\frac{1}{2016}\right)\times\left(1+\frac{1}{2017}\right)\)
\(A=\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times......\times\frac{2016}{2017}\times\frac{2018}{2017}\)
\(A=\frac{2018}{2}=1009\)
\(B=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.......+\frac{2}{43.45}\)
\(B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-......+\frac{1}{43}-\frac{1}{45}\)
\(B=\frac{1}{3}-\frac{1}{45}\)
\(B=\frac{14}{45}\)
2 \(\frac{2017}{2018}\times\frac{23}{47}+\frac{24}{2018}\times\frac{2017}{47}\)
\(=\frac{2017}{2018}\times\frac{23}{47}+\frac{24}{47}\times\frac{2017}{2018}\)
\(=\frac{2017}{2018}\times\left(\frac{23}{47}+\frac{24}{47}\right)\)
\(=\frac{2017}{2018}\times1\)
=\(\frac{2017}{2018}\)
bạn nào xem giải thế có đúng ko
\(a,\frac{5\cdot84\cdot105}{35\cdot50\cdot21}=\frac{1\cdot4\cdot3}{1\cdot10\cdot1}=\frac{1\cdot2\cdot3}{1\cdot5\cdot1}=\frac{6}{5}\)
\(b,\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{8}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{7}{8}\)
\(=\frac{1}{8}\)
a)
\(\frac{5\cdot84\cdot105}{35\cdot50\cdot21}\)
\(=\frac{5\cdot7\cdot2\cdot6\cdot35\cdot3}{35\cdot5\cdot10\cdot3\cdot7}\)
\(=\frac{6}{5}\)