Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)Ta có: \(\frac{1313}{1515}< \frac{1313}{1428}< \frac{1326}{1428}\Rightarrow\frac{1313}{1515}< \frac{1326}{1428}\)
b)Ta có: \(1-\frac{119}{120}=\frac{1}{120}< 1-\frac{118}{119}=\frac{1}{119}\Rightarrow\frac{119}{120}>\frac{118}{119}\)
c)Ta có: \(\frac{222}{555}< \frac{222}{444}< \frac{333}{444}\Rightarrow\frac{222}{555}< \frac{333}{444}\)
\(\frac{9}{11}\)< \(\frac{13}{15}\)
\(\frac{19}{15}\)< \(\frac{15}{11}\)
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 :
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{101.103}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{101.103}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{103}\right)\)
\(=\frac{1}{2}.\frac{102}{103}=\frac{51}{103}\)
\(\frac{29}{32}=\frac{29x3}{32x3}=\frac{87}{96};\frac{34}{48}=\frac{34x2}{48x2}=\frac{68}{96}\)
Vậy \(\frac{87}{96}>\frac{68}{96}\) Nên \(\frac{29}{32}>\frac{34}{48}\)
Ta có ; \(\frac{47}{15}=3\frac{2}{15};\frac{65}{21}=3\frac{2}{21}\)
Vì \(\frac{2}{15}>\frac{2}{21}\) Nên \(\frac{47}{15}>\frac{65}{21}\)
a) Ta có : \(\frac{2010}{2011}>\frac{2010}{2011+2012}\)
\(\frac{2011}{2012}>\frac{2011}{2011+2012}\)
Nên \(\frac{2010}{2011}+\frac{2011}{2012}>\frac{2010+2011}{2011+2012}\)=> M > N
b) P = \(\frac{2011.2012-2}{2010.2011+4020}=\frac{2011.\left(2010+2\right)-2}{2010.2011+4020}=\frac{2011.2010+2011.2-2}{2010.2011+4020}=\)\(\frac{2011.2010+4020}{2010.2011+4020}=1\)
Nên P = 1
câu b sửa lại:\(P=\frac{2011.2012-2}{2010.2011+4020}=\frac{2011.2010+4022-2}{2010.2011+4020}=\frac{2010.2011+4020}{2010.2011+4020}=1\)
a,\(\frac{7}{5}>\frac{5}{7}\)
b,\(\frac{14}{16}=\frac{24}{21}\)
a)
\(\frac{7}{5}>1\) ; \(\frac{5}{7}< 1\)
Nên \(\frac{7}{5}>\frac{5}{7}\)
b)
\(\frac{14}{16}< 1\) ; \(\frac{24}{21}>1\)
Nên \(\frac{14}{16}< \frac{24}{21}\)
a ,Ta có : 14/25 < 15/25 = 3/5
Ta có : 1 - 3/5 = 2/5
1 - 5/7 = 2/7
Nên 2/5 > 2/7
Vậy 14/25 < 5/7
\(a,\)Ta có:
\(1-\frac{167}{168}=\frac{1}{168}\)
\(1-\frac{176}{177}=\frac{1}{177}\)
Vì \(\frac{1}{168}>\frac{1}{177}\)
\(\Rightarrow\frac{167}{168}< \frac{176}{177}\)
A. 167/168 < 176/177
B. 169/168 < 178/177