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Ta có : \(3^{75}=3^{3.25}=\left(3^3\right)^{25}=27^{25}\)
\(2^{100}=2^{4.25}=\left(2^4\right)^{25}=16^{25}\)
Vì \(27>16\)
\(\Rightarrow\)\(27^{25}>16^{25}\)
\(\Rightarrow\)\(3^{75}>2^{100}\)
Vậy \(3^{75}>2^{100}\)
Tk nha ! Happy ♡♡♡
Ta có :
\(2^{100}=\left(2^4\right)^{25}=16^{25}\)
\(3^{75}=\left(3^3\right)^{25}=27^{25}\)
Có \(27>16\)
\(\Rightarrow\)\(27^{25}>16^{25}\)
Hay \(3^{75}>2^{100}\)
Ta có : \(10^{30}\)\(=\left(10^3\right)^{10}\)\(=1000^{10}\)
& \(2^{100}\)\(=\left(2^{10}\right)^{10}\)\(=1024^{10}\)
Vì : \(1000^{10}< 1024^{10}\)
\(\Rightarrow2^{100}>10^{30}\)
Ta có :
\(10^3=\left(10^3\right)^{10}=1000^{10}\)
\(2^{100}=\left(2^{10}\right)^{10}=1024^{10}\)
Vì \(1000^{10}< 1024^{10}\)
\(\Rightarrow10^3< 2^{100}\)
a) ta có A=\(15^{120}:25^{60}=3^{120}.5^{120}:5^{120}=3^{120}=9^{60}\)
B=\(2^{45}.2^{15}.4^{60}=2^{60}.2^{120}=2^{180}=8^{60}\)
-> A<B
b) bạn chỉ cần tính từng cái ra là dc ý ,ak dễ lắm nếu bạn chăm chỉ
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) \(2^{100}=\left(2^2\right)^{50}\)
\(2^2=4< 5\)
\(2^{100}< 5^{50}\)
b) \(4^{30}=\left(4^3\right)^{10}\)
\(4^3=8^2\)
\(4^{30}=8^{20}\)
\(8^{20}=\left(8^2\right)^{10}\)
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 \(\frac{81^{30}}{80^{30}}>1\)\(\Rightarrow\frac{27^{50}}{240^{30}}>1\)\(\Rightarrow27^{50}>240^{30}\)
a) \(\left(x-\frac{1}{2}\right)^4=\frac{1}{81}\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^4=\left(\frac{1}{3}\right)^4\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{1}{3}\\x-\frac{1}{2}=\frac{-1}{3}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=\frac{1}{6}\end{cases}}\)
Vậy ...
Ta có : 8175 = (813)25 = 53144125
30100 = (304)25 = 810 00025
Mà : 53144125 < 810 00025
Nên : 8175 < 30100
\(81^{75}< 30^{100}\)