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(4872 . 3478. 4876 ) . (1/2 . 4 -2)
=(4872 . 3478. 4876 ). ( 2 - 2 )
= (4872 . 3478. 4876 ) .0
=0
còn câu dưới mình ko biết
a, \(\frac{6}{7}.\frac{16}{15}.\frac{7}{6}.\frac{21}{32}=\frac{6}{7}.\frac{7}{6}.\frac{16}{15}.\frac{21}{32}\)=\(1.\frac{16}{15}.\frac{21}{32}=\frac{7}{5.2}=\frac{7}{10}\)
Phần b T2
c,\(\frac{7}{4}.\frac{11}{21}+\frac{11}{21}.\frac{5}{4}=\frac{11}{21}.\left(\frac{7}{4}+\frac{5}{4}\right)\)=\(\frac{11}{21}.3=\frac{11}{7}\)
a.\(\frac{3\cdot4\cdot7}{12\cdot8\cdot9}\)= \(\frac{3\cdot4\cdot7}{3\cdot4\cdot8\cdot9}\)= \(\frac{7}{72}\)
b. \(\frac{4\cdot5\cdot6}{12\cdot10\cdot8}\)= \(\frac{4\cdot5\cdot2\cdot3}{3\cdot4\cdot5\cdot2\cdot8}\)= \(\frac{1}{8}\)
c.\(\frac{5\cdot6\cdot7}{12\cdot14\cdot15}\)= \(\frac{5\cdot6\cdot7}{2\cdot6\cdot2\cdot7\cdot3\cdot5}\)= \(\frac{1}{12}\)
a, \(\frac{3.4.7}{12.8.9}\)= \(\frac{3.4.7}{3.4.8.9}\)= \(\frac{7}{72}\)
b, \(\frac{4.5.6}{12.10.8}\)= \(\frac{4.5.6}{3.4.2.5.8}\)= \(\frac{1}{8}\)
c, \(\frac{5.6.7}{12.14.15}\)= \(\frac{5.6.7}{2.6.2.7.3.5}\)= \(\frac{1}{12}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
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 :
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{101.103}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{101.103}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{103}\right)\)
\(=\frac{1}{2}.\frac{102}{103}=\frac{51}{103}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{1000\cdot1001}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1000}-\frac{1}{1001}\)
\(=1-\frac{1}{1001}\)
\(=\frac{1000}{1001}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{101}-\frac{1}{102}\)
\(=1-\frac{1}{102}\)
\(=\frac{101}{102}\)
1/1.2 + 1/2.3 + 1/3.4 + ... + 1/101.102
Đặt A = 1/1.2 +1/2.3 + 1/3.4 + ... + 1/101.102
A = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/101 - 1/102
A = 1/1 - 1/02
A = 101/102
Vậy A = 101/102
\(\frac{6\cdot7\cdot12}{18\cdot4\cdot7}\)
\(=\frac{6\cdot7\cdot4\cdot3}{6\cdot3\cdot4\cdot7}=1\)
\(\frac{15\cdot7\cdot9}{77\cdot3\cdot18}=\frac{5\cdot3\cdot7\cdot9}{11\cdot7\cdot3\cdot9\cdot2}=\frac{5}{11\cdot2}=\frac{5}{22}\)
chúc hok tốt
Trả lời
\(\frac{6.7.12}{18.4.7}=\frac{6.7.3.4}{6.3.4.7}=1\)
\(\frac{15.7.9}{77.3.18}=\frac{5.3.7.9}{11.7.3.2.9}=\frac{5}{22}\)
Học tốt