so sánh (-1/4)^8 và (1/8)^5
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\frac{1}{4}\right)^8=\left[\left(\frac{1}{4}\right)^2\right]^4=\left(\frac{1}{8}\right)^4\)
Vì \(\left(\frac{1}{8}\right)^4< \left(\frac{1}{8}\right)^5\Rightarrow\left(\frac{1}{4}\right)^8< \left(\frac{1}{8}\right)^5\).
Vậy \(\left(\frac{1}{4}\right)^8< \left(\frac{1}{8}\right)^5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: 3/5>-19/5
b: 8/7<8/3
c: 3/4=15/20
2/5=8/20
mà 15>8
nên 3/4>2/5
d: -3/5=-18/30
-4/6=-20/30
mà -18>-20
nên -3/5>-4/6
a: 3/5>-19/5
b: 8/7<8/3
c: 3/4=15/20
2/5=8/20
mà 15>8
nên 3/4>2/5
d: -3/5=-18/30
-4/6=-20/30
mà -18>-20
nên -3/5>-4/6
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
có:1/4+1/5+1/6+1/7+...+1/9≤nhỏ hơn 1/6.6=1
1/10+1/11+...+1/15 nhỏ hơn1/5.5=1
⇒1/4+1/5+...+1/15nhỏ hơn1+1=2(đpcm)
ta có
\(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}< \dfrac{1}{4}.4\)
\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}< 1\)
và:
\(\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< \dfrac{1}{8}.8\)
\(\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< 1\)
\(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{15}< 1+1=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
-1/48<1/85
\(\left(-\dfrac{1}{4}\right)^8=\left(\dfrac{1}{4}\right)^8=\left(\dfrac{1}{2}\right)^{16}\)
\(\left(\dfrac{1}{8}\right)^5=\left(\dfrac{1}{2}\right)^{15}\)
mà 16>15
nên \(\left(-\dfrac{1}{4}\right)^8>\left(\dfrac{1}{8}\right)^5\)