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ta có: 1- 2013/2015 = 2/ 2015
1- 2015/2017 = 2/2017
vì 2/2015 > 2/2017 nên 2013/2015 > 2015/2017
TÍCH CHO MIK NHA
\(a.\)\(\frac{601}{1203}<\frac{601}{1202}=\frac{1}{2}=\frac{511}{1022}<\frac{511}{1020}\)
b.\(\frac{347}{600}<\frac{347}{599}<\frac{374}{599}\)
Ta có : \(1-\frac{2013}{2015}=\frac{2}{2015}\)
\(1-\frac{2015}{2017}=\frac{2}{2017}\)
Mà : \(\frac{2}{2015}>\frac{2}{2017}\)nên \(\frac{2013}{2015}< \frac{2015}{2017}\)
Ta có: \(1-\frac{2013}{2015}=\frac{2}{2015}\)
\(1-\frac{2015}{2017}=\frac{2}{2017}\)
Mà \(\frac{2}{2015}>\frac{2}{2017}\Rightarrow\frac{2013}{2015}< \frac{2015}{2017}\)
\(\frac{9}{11}\)< \(\frac{13}{15}\)
\(\frac{19}{15}\)< \(\frac{15}{11}\)
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)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,\(\frac{7}{5}>\frac{5}{7}\)
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) \(\frac{13}{19}+\frac{18}{19}+\frac{19}{19}=\left(\frac{13}{19}+\frac{17}{19}\right)+\frac{18}{19}=\frac{20}{19}+\frac{18}{19}=\frac{48}{19}\)
b) \(\frac{3}{5}+\frac{3}{16}+\frac{13}{16}=\left(\frac{3}{16}+\frac{13}{16}\right)+\frac{3}{5}=1+\frac{3}{5}=\frac{5}{5}+\frac{3}{5}=\frac{8}{5}\)
a) \(\frac{13}{19}+\frac{18}{19}+\frac{17}{19}\)
= \(\frac{31}{19}+\frac{17}{19}\)
= \(\frac{48}{19}.\)
b) \(\frac{3}{5}+\frac{3}{16}+\frac{13}{16}\)
= \(\frac{3}{5}+\left(\frac{3}{16}+\frac{13}{16}\right)\)
= \(\frac{3}{5}+\frac{16}{16}\)
= \(\frac{3}{5}+1\)
= \(\frac{3}{5}.\)
Bài a:
a) \(1-\frac{203}{205}=\frac{2}{205}< 1-\frac{113}{115}=\frac{2}{115}\) => \(\frac{203}{205}>\frac{113}{115}\)
b) \(\frac{41}{70}>\frac{41}{73}>\frac{39}{73}\)