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a) \(2^{27}=\left(2^3\right)^9=8^9\)
\(3^{18}=\left(3^2\right)^9=9^9\)
b) ta có : \(2^{27}=\left(2^3\right)^9=8^9và3^{18}=\left(3^2\right)^9=9^9\)
vì : \(8^9< 9^9\)
nên: \(2^{27}< 3^{18}\)
a) \(2^{27}=\left(2^3\right)^9=8^9;3^{18}=\left(3^2\right)^9=9^9\)
b) Vì 89 < 99 => 227 < 318 nên 318 là số lớn hơn
a) Ta có:
b) Vì 8< 9 nên \(8^9< 9^9\)
Vậy theo câu a, ta được \(3^{18}< \) \(2^{27}\)
a) \(2^{27}=2^{3.9}=8^9\)
\(3^{18}=3^{2.9}=9^9\)
b) Vì \(8^9< 9^9\) nên \(2^{27}< 3^{18}\)
\(2^6\)\(0,5^2\)\(\left(\frac{1}{2}\right)^4\)\(\left(\frac{1}{2}\right)^8\)\(\left(\frac{11}{12}\right)^2\)
a)
\(10^8.2^8=\left(10.2\right)^8=20^8\)
b)
\(10^8:2^8=\left(10:2\right)^8=5^8\)
c)
\(25^4.2^8=\left(5^2\right)^4.2^8=5^8.2^8=\left(2.5\right)^8=10^8\)
d)
\(15^8.9^4=15^8.\left(3^2\right)^4=15^4.3^4=45^4\)
e)
\(27^2:25^3=\left(3^3\right)^2:\left(5^2\right)^3=3^6:5^6=\left(\frac{3}{5}\right)^6\)
a) \(=\left(\frac{-1}{5}^3\right)^{100}va\left(\frac{-1}{3}^5\right)^{100}\)
\(=\left(\frac{-1}{125}\right)^{100}va\left(\frac{-1}{243}\right)^{100}\)
Mà \(\frac{-1}{125}>\frac{-1}{243}\)
\(\Rightarrow\left(\frac{-1}{5}\right)^{300}>\left(\frac{-1}{3}\right)^{500}\)
b)\(2^{27}=8^9;3^{18}=9^9\)
Bài 1:
a, 2225 = (23)75 = 875
3150 = (32)75 = 975
Vì 875 < 975 nên 2225 < 3150
b, 212 = (24)3 = 163 ; 418 = (42)9 = 169
Bài 2:
a, 3300 = (33)100 = 27100
5200 = (52)100 = 25100
Vì 27100 > 25100 nên 3300 > 5200
b, Do \(\hept{\begin{cases}\left(x-3\right)^2\ge0\\\left|y^2-25\right|\ge0\end{cases}\forall x,y\Rightarrow\left(x-3\right)^2+\left|y^2-25\right|\ge0}\) (1)
Mà \(\left(x-3\right)^2+\left|y^2-25\right|=0\) (2)
Từ (1) và (2) => \(\hept{\begin{cases}\left(x-3\right)^2=0\\\left|y^2-25\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=3\\y=5\end{cases}}\)
Bài 3:
2x = -3y = 4z
=> \(\frac{2x}{12}=\frac{-3y}{12}=\frac{4z}{12}\)
=> \(\frac{x}{6}=\frac{-y}{4}=\frac{z}{3}\)
=> \(\frac{x}{6}=\frac{-2y}{8}=\frac{3z}{9}=\frac{x-2y-3z}{6+8-9}=\frac{30}{5}=6\)
=> x = 36, y = -24, z = 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 \(\frac{81^{30}}{80^{30}}>1\)\(\Rightarrow\frac{27^{50}}{240^{30}}>1\)\(\Rightarrow27^{50}>240^{30}\)
giúp mình nhé
b: \(2^{27}=8^9\)
\(3^{18}=9^9\)
mà 8<9
nên \(2^{27}< 3^{18}\)