so sánh \(\frac{1-a}{a}và0\)
b\(\frac{4}{9}và\frac{6+9}{6.9}và\frac{2}{3}\)
c\(\frac{18}{91}và\frac{23}{114}\)
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b/\(\frac{10^9+1}{10^{9+1}+1}\)=\(\frac{10^9+1}{10.10^9+1}\)=\(\frac{1}{10\text{}}\)
\(\frac{10^{10}+1}{10^{10+1}+1}\)=\(\frac{10^{10}+1}{10+10^{10}+1}\)=\(\frac{1}{10}\)
Vì \(\frac{1}{10}\)=\(\frac{1}{10}\)=>bằng nhau
ta co:
-18/91 va 23/-114
vi -18/91=-18:91 mot so am chia cho so duong
con 23/-114=23:114 mot so duong chia cho so am
nen -18/91<23/-114
quy đồng mẫu
-18 x 114=-2052
-23 x 91=-2093
vậy -18/91 > -23/114
a) Ta có :
\(\frac{18}{91}< \frac{18}{90}=\frac{1}{5}=\frac{23}{115}< \frac{23}{114}\)
\(\Rightarrow\frac{18}{91}< \frac{23}{114}\)
b) Ta có :
\(\frac{21}{52}=\frac{210}{520}=1-\frac{310}{520}\)
\(\frac{213}{523}=1-\frac{310}{523}\)
Mà \(1-\frac{310}{520}< 1-\frac{310}{523}\)
\(\Rightarrow\frac{21}{52}< \frac{213}{523}\)
c) Ta có : \(\frac{1313}{9191}=\frac{13}{91}=\frac{1}{7}=\frac{11}{77};\frac{1111}{7373}=\frac{11}{73}\)
Mà \(\frac{11}{77}< \frac{11}{73}\)nên \(\frac{1313}{9191}< \frac{1111}{7373}\)
d) Ta có :
\(\frac{n}{n+1}=\frac{n+1-1}{n+1}=1-\frac{1}{n+1}\)
\(\frac{n+2}{n+3}=\frac{n+3-1}{n+3}=1-\frac{1}{n+3}\)
Mà \(1-\frac{1}{n+1}< 1-\frac{1}{n+3}\)nên \(\frac{n}{n+1}< \frac{n+2}{n+3}\)
a) Ta có : \(\frac{18}{91}< \frac{18}{90}=\frac{1}{5}< \frac{23}{115}< \frac{23}{114}\)
\(\Rightarrow\) \(\frac{18}{91}< \frac{23}{114}\)
Vậy \(\frac{18}{91}< \frac{23}{114}\)
b) Ta có : \(\frac{21}{52}< \frac{21}{56}=\frac{3}{8}< \frac{213}{568}< \frac{213}{523}\)
\(\Rightarrow\) \(\frac{21}{52}< \frac{213}{523}\)
Vậy \(\frac{21}{52}< \frac{213}{523}\)
c) Ta có : \(\frac{1313}{9191}=\frac{1313:1313}{9191:1313}=\frac{1}{7}\)
\(\frac{1111}{7373}=\frac{1111:101}{7373:101}=\frac{11}{73}\)
Lại có : \(\frac{1}{7}< \frac{11}{77}< \frac{11}{73}\)
\(\Rightarrow\) \(\frac{1313}{9191}< \frac{1111}{7373}\)
Vậy \(\frac{1313}{9191}< \frac{1111}{7373}\)
d) Ta có : \(1-\frac{n}{n+1}=\frac{n+1}{n+1}-\frac{n}{n+1}=\frac{1}{n+1}\)
\(1-\frac{n+2}{n+3}=\frac{n+3}{n+3}-\frac{n+2}{n+3}=\frac{1}{n+3}\)
Vì \(n+1< n+3\)
\(\Rightarrow\)\(\frac{1}{n+1}>\frac{1}{n+3}\)
\(\Rightarrow\) \(\frac{n}{n+1}< \frac{n+2}{n+3}\)
Vậy \(\frac{n}{n+1}< \frac{n+2}{n+3}\)
Chúc m.n hok tốt ♡❤️
a)\(\frac{18}{91}\)< \(\frac{23}{114}\) ; b) \(\frac{1313}{9191}\) < \(\frac{1111}{7373}\)
a)\(\frac{18}{91}\)\(< \)\(\frac{23}{114}\)
b)\(\frac{1313}{9191}\)\(< \)\(\frac{1111}{7373}\)
Ta có: \(-\frac{18}{91}>-\frac{1}{5}=-\frac{18}{90}\)
\(-\frac{23}{114}-\frac{1}{5}>-\frac{23}{114}\)
Ta thấy
\(\frac{-18}{91}>\frac{-18}{114}\) (1)
\(\frac{-18}{114}>\frac{-23}{114}\) (2)
Từ (1) và (2) => \(\frac{-18}{91}>\frac{-23}{114}\)