\(\dfrac{2}{4} và\)\(\dfrac{3}{4} \)
so sánh
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: -3/100=-9/300; -2/3=-200/300
=>-3/100>-2/3
b: -3/5=-9/15
-2/3=-10/15
=>-3/5>-2/3
c: -5/4<-1<-3/8
d: -2/3=-8/12; -3/4=-9/12
=>-2/3>-3/4
e: -267/268>-1
-1>-1347/1343
=>-267/268>-1347/1343
![](https://rs.olm.vn/images/avt/0.png?1311)
M=1/4(4/1*5+8/5*13+...+16/25*41)
=1/4(1-1/5+1/5-1/13+...+1/25-1/41)
=40/41*1/4=10/41
\(N=\dfrac{1}{3}\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{16}+...+\dfrac{1}{43}-\dfrac{1}{61}\right)=\dfrac{1}{3}\cdot\dfrac{60}{61}=\dfrac{20}{61}\)
=>M<N
![](https://rs.olm.vn/images/avt/0.png?1311)
Cách 1:
\(\dfrac{3}{4}=\dfrac{9}{12}\)
\(\dfrac{4}{3}=\dfrac{16}{12}\)
Do đó \(\dfrac{3}{4}< \dfrac{4}{3}\)
Cách 2:
\(\dfrac{3}{4}< 1\)
\(1< \dfrac{4}{3}\)
Do đó \(\dfrac{3}{4}< \dfrac{4}{3}\)
\(-------\)
Cách 1:
\(\dfrac{11}{8}=\dfrac{55}{40}\)
\(\dfrac{7}{10}=\dfrac{28}{40}\)
Do đó \(\dfrac{11}{8}>\dfrac{7}{10}\)
Cách 2:
\(\dfrac{11}{8}>1\)
\(1>\dfrac{7}{10}\)
Do đó \(\dfrac{11}{8}>\dfrac{7}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$\frac{n+3}{n+4}=\frac{(n+4)-1}{n+4}=1-\frac{1}{n+4}$
$\frac{n+1}{n+2}=\frac{(n+2)-1}{n+2}=1-\frac{1}{n+2}$
Vì $n+4> n+2$ nên $\frac{1}{n+4}< \frac{1}{n+2}$
Suy ra $1-\frac{1}{n+4}> 1-\frac{1}{n+2}$
Hay $\frac{n+3}{n+4}> \frac{n+1}{n+2}$
-------------------------
$\frac{n-1}{n+4}< \frac{n-1}{n+2}=\frac{(n+2)-3}{n+2}=1-\frac{3}{n+2}$
$<1-\frac{n+3}=\frac{n}{n+3}$
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\(\dfrac{2}{4}< \dfrac{3}{4}\)