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a,\(27^{11}=\left(3^3\right)^{11}=3^{33}\)
\(81^8=\left(3^4\right)^8=3^{32}\)
Vì \(3^{33}>3^{32}\Rightarrow27^{11}>81^8\)
b,\(3^{150}=\left(3^2\right)^{75}=9^{75}\\ 2^{225}=\left(2^3\right)^{75}=8^{75}\)
Vì \(9^{75}>8^{75}\Rightarrow3^{150}>2^{225}\)
c,Ta có :\(\dfrac{27^{50}}{240^{30}}\)
=\(\dfrac{27^{30}.27^{20}}{240^{30}}\)
=\(\dfrac{3^{30}.3^{30}.3^{30}.3^{20}.3^{20}.3^{20}}{3^{30}.80^{30}}\)
=\(\dfrac{3^{120}}{80^{30}}\)=\(\dfrac{\left(3^4\right)^{30}}{80^{30}}\)=\(\dfrac{81^{30}}{80^{30}}\)
\(\rightarrow\)\(\dfrac{81^{30}}{80^{30}}>1\rightarrow\dfrac{27^{50}}{240^{30}}>1\Rightarrow27^{50}>240^{30}\)
a)ta có 2711 và 818
= (33)11 và (34)8
=333 và 332
do 33>32
=>333>332hay2711>818
a) ta có: 2225 = (23)75 = 875
3150 = (32)75 = 975 > 875
=> ...
b) ta có: 291 > 275 = (23)25 = 825 > 325
=> ...
c) ta có: 278 = (33)8 = 324
814 = (34)4 = 316 < 324
=>...
d)ta có: 2332 < 2333 = (23)111 = 8111
3223 > 3222 = (32)111 = 9111 > 8111
=>...
e)C1: ta có: 92000 = (32)2000 = 34000
C2: ta có: 34000 = (32)2000 = 92000
\(2^{225}=\left(2^3\right)^{75}=8^{75}\)
\(3^{150}=\left(3^2\right)^{75}=9^{75}\)
\(8< 9=>....\)
a) Ta có :
\(\hept{\begin{cases}27^{11}=\left(3^3\right)^{11}=3^{33}\\81^8=\left(3^4\right)^8=3^{32}\end{cases}}\)
Vì 333 > 332
=> 2711 > 818
b) Ta có:
\(\hept{\begin{cases}2^{225}=\left(2^3\right)^{75}=8^{75}\\3^{150}=\left(3^2\right)^{75}=9^{75}\end{cases}}\)
Vì 875 < 975
=> 2225 < 3150
Thôi còn lại bn tự làm nốt nha . Nhìn mà nản !!
a) \(\hept{\begin{cases}27^{11}=\left(3^3\right)^{11}=3^{33}\\81^8=\left(3^4\right)^8=3^{32}\end{cases}}\)
333 > 332 => 2711 > 818
b) \(\hept{\begin{cases}2^{225}=\left(2^3\right)^{75}=8^{75}\\3^{150}=\left(3^2\right)^{75}=9^{75}\end{cases}}\)
875 < 975 => 2225 < 3150
c) \(\hept{\begin{cases}2^{500}=\left(2^5\right)^{100}=32^{100}\\5^{200}=\left(5^2\right)^{100}=25^{100}\end{cases}}\)
32100 > 25100 => 2500 > 5200
d) \(\hept{\begin{cases}625^5=\left(5^4\right)^5=5^{20}\\125^7=\left(5^3\right)^7=5^{21}\end{cases}}\)
520 < 521 => 6255 < 1257
e) \(\hept{\begin{cases}5^{100}=\left(5^4\right)^{25}=625^{25}\\8^{75}=\left(8^3\right)^{25}=512^{25}\end{cases}}\)
62525 > 51225 => 5100 > 875
f) \(2^{16}=2^3\cdot2^{13}=8\cdot2^{13}\)
7 < 8 => 7.213 < 8.213 => 7.213 < 216
g) Ta có \(\frac{27^{50}}{240^{30}}=\frac{\left(3^3\right)^{50}}{3^{30}\cdot80^{30}}=\frac{3^{150}}{3^{30}\cdot80^{30}}=\frac{3^{120}}{80^{30}}=\frac{\left(3^4\right)^{30}}{80^{30}}=\frac{81^{30}}{80^{30}}\)
Vì 8130 > 8030 => 8130/8030 > 1 => 2750/24030 > 1 => 2750 > 24030
h) Ta có \(\hept{\begin{cases}63^9< 64^9=\left(2^6\right)^9=2^{54}\left(1\right)\\16^{14}=\left(2^4\right)^{14}=2^{56}< 17^{14}\left(2\right)\end{cases}}\)
Từ (1) và (2) => 639 < 254 < 256 < 1714
=> 639 < 1714
a, 2300 = (23)100 = 8100
3200 = (32)100 = 9100
Vì 8100 < 9100
=> 2300 < 3200
b, 220 = (25)4 = 324
312 = (33)4 = 274
Vì 324 > 274
=> 220 > 312
c, 2225 = (23)75 = 875
3150 = (32)75 = 975
Vì 875 < 975
=> 2225 < 3150
d, 2115 = (3.7)15 = 315.715
275.498 = (33)5.(72)8 = 315.716
Vì 315.715 < 315.716
=> 2115 < 275.498
a) Ta có: 1,(375) = 1,375375375…
\(1\frac{3}{8}\) = 1,375
Vì 1,375375... > 1,375 nên 1,(375) > \(1\frac{3}{8}\)
b) Ta có: -1,(27) = -1,272727…
Vì 1,272727… > 1,272 nên - 1,272727 < -1,272 hay – 1,(27) < -1,272
vì 1/27^11 = 1/(3^3)^11 = 1/3^33
1/81^8= 1/(9^2)^8 = 1/9^16 = 1/(3^2)^16 = 1/3^32
=> <
vì 1/3^99= 1/(3)^33.3=1/99^3 = 1/ (33.3)^3 = 1/33^9
1/11^21=1/(11)^3.7=1/33^7
=> " > "
nhớ ****
b/ Ta có: 291>290=(25)18=3218>2518=(52)18=536>535 => 291>535
c/ Ta có: 2225=(23)75=875
3150=(32)75=975
Vì 875<975 nên 2225<3150
a)Ta có: 2^27=(2^3)^9=8^9
3^18=(3^2)^9=9^9
Vì 8^9 <9^9
2^27<3^18
d)Ta có :27^7=(3^3)^7=3^21
9^12=(3^2)^12=3^24
Vì 3^21<3^24
27^7<9^12
Ta có: \(99^{20}=\left(99^2\right)^{10}=9801^{10}\)
Vì \(9801< 9999\)nên \(9801^{10}=9999^{10}\)
Vậy \(99^{20}< 9999^{10}\)
a, 2^27 = 2^3.9 = 8^9
3^18 = 3^2.9 = 9^9
vì 8<9 => 8^9 < 9^9 => 2^27 < 3^18
a)27^11=(3^3)^11=3^33
81^8=(3^4)8=3^32
vì 3^33>3^32 nên 27^11>81^8
b)ko biết làm chỉ biết 3^150>2^225
c)27^50=27^5x10=(27^5)^10=14348907^10
240^30=240^3x10=(240^3)^10=13824000^10
suy ra 27^50>240^30
a) Ta có: \(27^{11}=\left(3^3\right)^{^{11}}=3^{3.11}=3^{33}\)
\(81^8=\left(3^4\right)^{^8}=3^{4.8}=3^{32}\)
Vì \(3^{33}>3^{32}\)
nên \(27^{11}>81^8\)
b) Ta có: \(3^{150}=3^{2.75}=\left(3^2\right)^{^{75}}=9^{75}\)
\(2^{225}=2^{3.75}=\left(2^3\right)^{^{75}}=8^{75}\)
vì \(9^{75}>8^{75}\)
nên \(3^{150}>2^{225}\)
c) Ta có:
\(\frac{27^{50}}{240^{30}}=\frac{27^{30}.27^{20}}{240^{30}}=\frac{3^{30}.3^{30}.3^{30}.3^{20}.3^{20}.2^{20}}{3^{30}.80^{30}}\)
\(=\frac{3^{120}}{80^{30}}=\frac{\left(3^4\right)^{^{30}}}{80^{30}}=\frac{81^{30}}{80^{30}}\)
Vì \(\frac{81^{30}}{80^{30}}>1\)\(\Rightarrow\frac{27^{50}}{240^{30}}>1\)\(\Rightarrow27^{50}>240^{30}\)