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\(\dfrac{5}{-6}=\dfrac{-55}{66};\dfrac{-10}{11}=\dfrac{-60}{66}\Rightarrow\dfrac{-55}{66}>\dfrac{-60}{66}\Rightarrow\dfrac{5}{-6}>\dfrac{-10}{11}\\ \dfrac{-3}{20}=\dfrac{-45}{300};\dfrac{2}{-15}=\dfrac{-40}{300}\Rightarrow\dfrac{-45}{300}< \dfrac{-40}{300}\Rightarrow\dfrac{-3}{20}< \dfrac{2}{-15}\\ -0,305>-0,36\)
\(^\circ\) \(\dfrac{5}{-6}\) và \(\dfrac{-10}{11}\)
Ta có \(:\)
\(\dfrac{5}{-6} = \dfrac{ 5 . 11 }{ -6 . 11 } = \dfrac{ 55 }{ -66} \)
\(\dfrac{-10}{11} = \dfrac{-10 . ( -6 )}{11.(-6)} = \dfrac{60}{-66}\)
Do \(55 < 60\)
\(=> \dfrac{55}{-66} > \dfrac{60}{-66}\)
Vậy \(\dfrac{55}{-66} > \dfrac{60}{-66}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{{ - 3}}{8} = \frac{{ - 3.3}}{{8.3}} = \frac{{ - 9}}{{24}}\)
Vì -9 < -5 nên \(\frac{{ - 9}}{{24}} < \frac{{ - 5}}{{24}}\)
Vậy \(\frac{{ - 3}}{8} < \frac{{ - 5}}{{24}}\).
b) Cách 1: \(\frac{{ - 2}}{{ - 5}} = \frac{2}{5}; \frac{3}{{ - 5}} = \frac{-3}{{5}}\)
Vì 2 > -3 nên \(\frac{2}{5} > \frac{-3}{{5}}\)
Vậy \(\frac{{ - 2}}{{ - 5}} > \frac{3}{{ - 5}}\).
Cách 2: \(\frac{{ - 2}}{{ - 5}} = \frac{2}{5} > 0\) mà \(\frac{3}{{ - 5}} < 0\)
\(\Rightarrow\) \(\frac{{ - 2}}{{ - 5}} > \frac{3}{{ - 5}}\).
c) \(\frac{{ - 3}}{{ - 10}} = \frac{3}{{10}} = \frac{{3.2}}{{10.2}} = \frac{6}{{20}}\)
\(\frac{{ - 7}}{{ - 20}} = \frac{7}{{20}}\)
Vì 6 < 7 nên \(\frac{6}{{20}} < \frac{7}{{20}}\) nên \(\frac{{ - 3}}{{ - 10}} < \frac{{ - 7}}{{ - 20}}\).
d) \(\frac{{ - 5}}{4} = \frac{{ - 5.5}}{{4.5}} = \frac{{ - 25}}{{20}}; \frac{{ 23}}{{-20}}=\frac{{-23}}{{20}} \)
Vì -25 < -23 nên \( \frac{{ - 25}}{{20}} < \frac{{-23}}{{20}} \)
Vậy \(\frac{{ - 5}}{4} < \frac{{23}}{{ - 20}}\).
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f: 11^1979<11^1980=1331^660
37^1320=(37^2)^660=1369^660
1331<1369
=>1331^660<1369^660
=>11^1980<37^1320
=>11^1979<37^1320
g: 10^10=2^10*5^10
48*50^5=2^4*3*2^5*5^10=2^9*3*5^10
2^10<2^9*3
=>2^10*5^10<2^9*3*5^10
=>10^10<48*50^5
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Ta có: 5^30 = (5^3)^10= 125^ 10 > (-10^2)^10= 100^10
b, ta có: 21^12= ( 21^3)^4 > 54^4
c, Ta có: (1/16)^10 = 1/16^10
(1/2)^50= 1/2^50
Lại có: 16^10=(2^4)^10= 2^40 < 2^50 nên (1/6)^10> (1/2)^50
![](https://rs.olm.vn/images/avt/0.png?1311)
a)9^20 và 27^13
9^20=(3^2)^20=3^40
27^13=(3^3)^13=3^39
vì 3^40 > 3^39 =>9^20>27^13
b)10^30 và 2^100
10^30=(10^3)^10=30^10
2^100=(2^10)^10=20^10
vì 30^10>20^0 => 10^30>2^100
c)125^5 và 25^7
125^5=(5^3)^5=5^15
25^7=(5^2)^7=5^14
vì 5^15>5^14 =>125^5>25^7
Ta có :
a) \(9^{20}=\left(3^2\right)^{20}=3^{40};27^{13}=\left(3^3\right)^{13}=3^{39}\)
Vì \(3^{40}>3^{39}\Rightarrow9^{20}>27^{13}\)
Vậy \(9^{20}>27^{13}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
8:
\(A=\dfrac{20^{10}-1+2}{20^{10}-1}=1+\dfrac{2}{20^{10}-1}\)
\(B=\dfrac{20^{10}-3+2}{20^{10}-3}=1+\dfrac{2}{20^{10}-3}\)
mà 20^10-1>20^10-3
nên A<B
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Đặt A = 1 + 2 + 22 + ... + 22008 (1)
=> 2A = 2 + 22 + 23 + ... + 22009 (2)
Lấy (2) trừ (1) theo vế ta có :
2A - A = (2 + 22 + 23 + ... + 22009) - (1 + 2 + 22 + ... + 22008)
A = 22009 - 1
Khi đó B = \(\frac{2^{2009}-1}{1-2^{2009}}=\frac{2^{2009}-1}{-\left(2^{2009}-1\right)}=-1\)
b) Ta có A = \(\frac{20^{10}+1}{20^{10}-1}\)
=> A - 1 = \(\frac{20^{10}+1-20^{10}+1}{20^{10}}=\frac{2}{20^{10}}\)
Lại có B = \(\frac{20^{10}-1}{20^{10}-3}\)
=> B - 1 = \(\frac{20^{10}-1-20^{10}+3}{20^{10}-3}=\frac{2}{2^{10}-3}\)
Vì \(\frac{2}{2^{10}}< \frac{2}{2^{10}-3}\)
=> A - 1 < B - 1
=> A < B
a) \(B=\frac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}\)
Đặt \(Q=1+2+2^2+...+2^{2008}\)
\(2Q=2+2^2+2^3+...+2^{2009}\)
\(2Q-Q=2+2^2+2^3+...+2^{2009}-1-2-2^2-...-2^{2008}\)
\(\Rightarrow Q=2^{2009}-1\)
Ta thấy \(Q\) là số đối của \(2^{2009}-1\)
\(\Rightarrow B=-1\)
Vậy \(B=-1\).
b) Ta có: \(A=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\)
Ta lại có: \(B=\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3+2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)
Vì \(\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\) nên \(1+\frac{2}{20^{10}-1}< 1+\frac{2}{20^{10}-3}\)
\(\Rightarrow A< B\)
Vậy \(A< B\).
Ta có :
( 52 )10 = 52.10 = 520 < 620 = ( 2 . 3 )20 = 220 . 320
Vậy ( 52 )10 < 220 . 320
~~Học tốt~~