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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 ♡❤️
quy đồng các phân số sao cho chúng cùng mẫu là so sánh được
Ta có:
a)18/91=18:91=0,197802197
23/114=23:114=0,201754386
Mà:0,197802197<0,201754386 nên 18/91<23/114
b)21/52=21:52=0,403846153
213/523=213:523=0,407265774
Mà:0,403846153<0,407265774 nên 21/52<213/523
c)1313/9191=1313:9191=0,142857142
1111/7373=1111:7373=0,150684931
Mà:0,142857142<0,150684931 nên 1313/9191<1111/7373
^^^^!~~~
a)\(\frac{18}{91}\)< \(\frac{23}{114}\) ; b) \(\frac{1313}{9191}\) < \(\frac{1111}{7373}\)
a)\(\frac{18}{91}\)\(< \)\(\frac{23}{114}\)
b)\(\frac{1313}{9191}\)\(< \)\(\frac{1111}{7373}\)
a). n/n+1 < n+2/n+3
b). n/n+3 > n−1/n+4
c). n/2n+1 < 3n+1/6n+3
k mk nha
\(\frac{n}{n+1}< 1\Rightarrow\frac{n}{n+1}< \frac{n+2}{n+1+2}=\frac{n+2}{n+3}\)
=>n/n+1<n+2/n+3
vậy........
b)\(\frac{n}{n+3}>\frac{n}{n+4}>\frac{n-1}{n+4}\Rightarrow\frac{n}{n+3}>\frac{n}{n+4}\)
vậy.....
c)\(\frac{n}{2n+1}=\frac{3n}{6n+3}< \frac{3n+1}{6n+3}\)
vậy.......
a) quy đồng : \(\frac{n}{2n+1}=\frac{3n}{6n+3}\)
Vì 3n < 3n + 1 => \(\frac{3n}{6n+3}< \frac{3n+1}{6n+3}\)hay \(\frac{n}{2n+1}< \frac{3n+1}{6n+3}\)
b) Ta có :
\(\frac{n}{n+1}=\frac{n+1-1}{n+1}=1-\frac{1}{n+1}\)
\(\frac{n+2}{n+3}=\frac{n+3-1}{n+3}=1-\frac{1}{n+3}\)
Vì \(\frac{1}{n+1}>\frac{1}{n+3}\)nên \(1-\frac{1}{n+1}< 1-\frac{1}{n+3}\)hay \(\frac{n}{n+1}< \frac{n+2}{n+3}\)
c) giả sử \(\frac{n}{n+3}< \frac{n-1}{n+4}\)
\(\Leftrightarrow\frac{n\left(n+4\right)}{\left(n+3\right)\left(n+4\right)}< \frac{\left(n-1\right)\left(n+3\right)}{\left(n+3\right)\left(n+4\right)}\)
\(\Rightarrow n^2+4n< n^2+2n-3\)
\(\Rightarrow2n< -3\)( vô lí )
Vậy \(\frac{n}{n+3}>\frac{n-1}{n+4}\)
