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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 :
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{101.103}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{101.103}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{103}\right)\)
\(=\frac{1}{2}.\frac{102}{103}=\frac{51}{103}\)
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: \(\frac{222}{555}< \frac{222}{444}< \frac{333}{444}\Rightarrow\frac{222}{555}< \frac{333}{444}\)
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
\(\frac{9}{11}\)< \(\frac{13}{15}\)
\(\frac{19}{15}\)< \(\frac{15}{11}\)
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 \(\frac{14}{16}< \frac{24}{21}\)
1)\(\frac{7}{8}>\frac{6}{7}>\frac{4}{5}>\frac{1}{2}>\frac{5}{16}\)
2)
a.\(\frac{3}{7}\)và\(\frac{5}{16}\)
Ta có :\(\frac{3}{7}=\frac{3\times5}{7\times5}=\frac{15}{35}\) \(\frac{5}{16}=\frac{5\times3}{16\times3}=\frac{15}{48}\)
\(\frac{15}{35}>\frac{15}{48}\Rightarrow\frac{3}{7}>\frac{5}{16}\)
b.làm tương tự như câu a nhé
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:\(P=\frac{2011.2012-2}{2010.2011+4020}=\frac{2011.2010+4022-2}{2010.2011+4020}=\frac{2010.2011+4020}{2010.2011+4020}=1\)
1)2007/2006 > 1
2006/2008 < 1
2007/2006 > 1 > 2006/2008
2007/2006 >2006 /2008
2)1313/1414 = 1313: 101/1414 : 101 =13/14
13/14 = 13/14 => 1313/1414= 13/14
3) 97/96 - 1 = 1/96
96/95 - 1 = 1/95
1/96<1/95 => 97/96 < 96/95
4) 2007/2006 - 1 = 1/2006
2005/2004 - 1 = 1/2004
1/2006 < 1/2004 => 2007/2006 < 2005/2004
5) 2007/2006 - 1 = 1/2006
2008/2007 - 1 = 1/2007
1/2006 > 1/2007 => 2007/2006 > 2008/2007
1) so sánh với 1
2) rút gọn rồi so sánh
3,4,5 ) dùng phần bù ( phân số có phần bù lớn hơn thì phân số đó lớn hơn )
\(a,\)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