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\(a,\frac{8}{9}< \frac{108}{109}\)
\(b,\frac{97}{100}< \frac{98}{99}\)
\(c,\frac{19}{18}>\frac{2017}{2016}\)
\(d,\frac{15}{16}>\frac{515}{616}\)
a)
\(\frac{64}{85}< \frac{64}{81}< \frac{73}{81}\)
=>\(\frac{64}{85}< \frac{73}{81}\)
b)
\(\frac{25}{26}=\frac{25.1010}{26.1010}=\frac{25250}{26260}\)
Ta có: \(1-\frac{25250}{26260}=\frac{1010}{26260}\)
\(1-\frac{25251}{26261}=\frac{1010}{26261}\)
Vì \(\frac{1010}{26260}>\frac{1010}{26261}\) nên \(\frac{25}{26}< \frac{25251}{26261}\)
a)\(\frac{64}{85}\)<\(\frac{64}{81}\)<\(\frac{73}{81}\)
b)\(\frac{25}{26}\)=\(\frac{25250}{26260}\)=\(1\)- \(\frac{1010}{26260}\)< \(1\)- \(\frac{1010}{26261}\)= \(\frac{25251}{26261}\)
a) \(\frac{8}{9}=1-\frac{1}{9}\)
\(\frac{108}{109}=1-\frac{1}{109}\)
Vì \(\frac{1}{9}>\frac{1}{109}\)
Nên \(1-\frac{1}{9}< 1-\frac{1}{109}\)
Vậy \(\frac{8}{9}< \frac{108}{109}\)
b)
\(\frac{97}{100}=\frac{97\cdot99}{100\cdot99}\)
\(\frac{98}{99}=\frac{98\cdot100}{99\cdot100}\)
\(\Rightarrow\frac{97}{100}< \frac{98}{99}\)
c)
\(\frac{19}{18}=1+\frac{1}{18}\)
\(\frac{2017}{2016}=1+\frac{1}{2016}\)
Vì \(\frac{1}{18}>\frac{1}{2016}\)
Vậy \(\frac{19}{18}>\frac{2017}{2016}\)
d)
\(\frac{133}{173}=\frac{130+3}{170+3}=\frac{13+0,3}{17+0,3}\)
Ta có :
\(\frac{a}{b}< \frac{a+x}{b+x}\forall a;b;x>0\)
Vậy \(\frac{13}{17}< \frac{133}{173}\)
\(a)\) Ta có :
\(\frac{51}{85}=\frac{3}{5}\)
\(\frac{58}{145}=\frac{2}{5}\)
Vì \(\frac{3}{5}>\frac{2}{5}\) nên \(\frac{51}{85}>\frac{58}{145}\)
Vậy \(\frac{51}{85}>\frac{58}{145}\)
\(b)\) Ta có :
\(\frac{69}{-230}=\frac{-3}{10}\)
\(\frac{-39}{143}=\frac{-3}{11}\)
Vì \(\frac{-3}{10}< \frac{-3}{11}\) nên \(\frac{69}{-230}< \frac{-39}{143}\)
Vậy \(\frac{69}{-230}< \frac{-39}{143}\)
\(c)\) Ta có :
\(1+\frac{-7}{41}=\frac{34}{41}\)
\(1+\frac{13}{-47}=\frac{34}{47}\)
Vì \(\frac{34}{41}>\frac{34}{47}\) nên \(1+\frac{-7}{41}>1+\frac{13}{-47}\) hay \(\frac{-7}{41}>\frac{13}{-47}\)
Vậy \(\frac{-7}{41}>\frac{13}{-47}\)
\(d)\) Ta có :
\(1-\frac{40}{49}=\frac{9}{49}\)
\(\frac{15}{21}=\frac{5}{7}=\frac{35}{49}< \frac{40}{49}\)
Vậy \(\frac{40}{49}>\frac{15}{21}\)
h) Ta có: \(\frac{n+1}{n+2}=1-\frac{1}{n+2}\)
\(\frac{n+3}{n+4}=\frac{1}{n+4}\)
Vì \(n+2< n+4\)\(\Rightarrow\frac{1}{n+2}>\frac{1}{n+4}\)
\(\Rightarrow1-\frac{1}{n+2}< 1-\frac{1}{n+4}\)\(\Rightarrow\frac{n+1}{n+2}< \frac{n+3}{n+4}\)
a) 13/57=13+16/57+16=29/73 ( Ghi nhớ SKG Toán 6)
-=> 13/57 < 29/73
b) 17/42 = 17-4/42-4 = 13/38
=> 17/42 > 13/38
c)7/41 = 7+6/41+6= 13/47
=> 7/41<13/47
e) \(\frac{15}{16}=\frac{15.1010}{16.1010}=\frac{15150}{16160}=1-\frac{1010}{16160}\)
\(\frac{15151}{16161}=1-\frac{1010}{16161}\)
Vì \(16160< 16161\)\(\Rightarrow\frac{1}{16160}>\frac{1}{16161}\)
\(\Rightarrow\frac{1010}{16160}>\frac{1010}{16161}\)\(\Rightarrow1-\frac{1010}{16160}< 1-\frac{1010}{16161}\)
hay \(\frac{15}{16}< \frac{15151}{16161}\)