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a) ta có: 27^4 = (3^3)^4 = 3^12 < 3^20
b) ta có: 25.5^31 = 5^2.5^31 = 5^33 < 5^34
c) ta có: 16^504 = (2^4)^504 = 2^2016
32^403 = (2^5)^403 = 2^2015 < 2^2016
=> 16^504 > 32^403
d) ta có: 5^301 > 5^300 = (5^3)^100 = 125^100
11^199 < 11^200 = (11^2)^100 = 121^100
=> 125^100 > 121^100
=> 5^301 > 11^199
a)ta có : 27^4=(3^3)^4=3^12<3^20
=>27^4<3^20
b)ta có :25*5^31= 5^2*5^31=5^33<5^34
=>5^34>25*5^31
c)ta có :16^504= (2^4)^504=2^2016
32^403=(2^5)^403=2^2015
=>2^2016>2^2015
=>16^504>32^403
d)5^301=125^100*5=121^100*5*4^100
11^199=121^99*11<121^100*5*4^100
=>5^301>11^199
1, A = 291 = 27.13 = (213)7 = 81927
B = 535 = 55.7 = (55)7 = 31257
Vì 3125 < 8192
=> 31257 < 81927
=> B < A
2.Ta có:
A=11+112+113+114+...+11199+11200.
11A=112+113+114+...+11199+11200+11201.
11A-A=11201-11.
10A=11201-11.
A=(11201-11):10
Quan sát 2 vế A và B thì ta thấy rõ ràng vế A<B hay B>A.
a,
\(-\frac{13}{38}=-1--\frac{25}{38}=-1+\frac{25}{38}\)
\(\frac{29}{-88}=-\frac{29}{88}=-1--\frac{59}{88}=-1+\frac{59}{88}\)
Vì \(\frac{25}{38}< \frac{59}{88}\Rightarrow-\frac{13}{38}< \frac{29}{-88}\)
b,
Ta có:
3301 > 3300 = [33]100 = 27100
5199 < 5200 = [52]100 = 25100
Mà 27100 > 25100 => 3301 > 5199
c,
\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left[2n+1\right]\left[2n+3\right]}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n+1}-\frac{1}{2n+3}\)
\(=1-\frac{1}{2n+3}< 1\)
Vậy P < 1
\(5^{199}=\left(5^{\frac{199}{301}}\right)^{301}\)
\(5^{\frac{199}{301}}< 3^1\)
\(\Leftrightarrow5^{199}< 3^{301}\)
cho 149 là 150
7150 = 73.50 = 34350
34550 = 3451.50 = 34550
Mà 34350 < 34550 Nên:
7149 <34550
Cho 199 là 200
3200 = 32.100 = 9100
11100 = 111.100 = 11100 Mà
9100 < 11100 Nên
3199 < 11100
7^149 < 7^150 = 343^50 < 345^50 => 7^149 < 345^50
3^199 < 3^200 = 9^100 < 11^100 => 3^199 <11^100
a, Ta có : \(\frac{13}{38}>\frac{13}{39}=\frac{1}{3}=\frac{29}{87}>\frac{29}{88}\)
\(\Rightarrow\frac{13}{38}>\frac{29}{88}\Rightarrow\frac{-13}{38}< \frac{29}{-88}\)
b, Ta có: \(3^{301}>3^{300}=\left(3^3\right)^{100}=27^{100}\left(1\right)\)
\(5^{199}< 5^{200}=\left(5^2\right)^{100}=25^{100}\left(2\right)\)
Do \(25^{100}< 27^{100}\Rightarrow5^{200}< 3^{300}\)\(\left(3\right)\)
Từ \(\left(1\right),\left(2\right),\left(3\right)\Rightarrow5^{199}< 5^{200}< 3^{300}< 3^{301}\Rightarrow5^{199}< 3^{301}\)
c, Ta có: \(\frac{10^{2018}+5}{10^{2018}-8}=\frac{10^{2018}-8+13}{10^{2018}-8}=1+\frac{13}{10^{2018}-8}\)
\(\frac{10^{2019}+5}{10^{2019}-8}=\frac{10^{2019}-8+13}{10^{2019}-8}=1+\frac{13}{10^{2019}-8}\)
Do \(\frac{13}{10^{2018}-8}>\frac{13}{10^{2019}-8}\Rightarrow1+\frac{13}{10^{2018}-8}>1+\frac{13}{10^{2019}-8}\Rightarrow\frac{10^{2018}+5}{10^{2018}-8}>\frac{10^{2019}+5}{10^{2019}-8}\)
a) \(3^{39}\) và \(11^{21}\)
\(\Rightarrow3^{39}=3^{13.3}=1594323^3\)
\(\Rightarrow11^{21}=11^{7.3}=194487171^3\)
Nên \(3^{39}< 11^{21}\)
b) \(199^{20}\) và \(2003^{15}\)
\(\Rightarrow199^{20}=199^{4.5}=1568239201^5\)
\(\Rightarrow2003^{15}=8036054027^5\)
Nên \(199^{20}< 2003^{15}\)