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a) ta có: (-32)9 = [(-2)5 ]9 = (-2)45 = - (2)45
(-16)13 = - [ 24 ]13 = - (2)52
=> ....
b) ta có: (-5)30 = 530 = (53)10 = 12510
(-3)50 = 350 = (35)10 = 24310
=> ....
c) ta có: (-32)9 = (-2)45 = (-2)13 . 232
(-18)13 = [(-2).32 ]13 = (-2)13 . 339
=> ....
d) ta có: \(\left(-\frac{1}{16}\right)=-\left(\frac{1}{2}\right)^4.\)
\(\left(-\frac{1}{2}\right)=-\left(\frac{1}{2}\right)^1< -\left(\frac{1}{2}\right)^4\)
a)Ta có:\(3^{30}=\left(3^3\right)^{10}=27^{10}\)
\(5^{20}=\left(5^2\right)^{10}=25^{10}\)
Vì \(27^{10}>25^{10}\Rightarrow3^{30}>5^{20}\)
Do 27>25 nên \(27^{10}>25^{10}\)\(hay\) \(3^{30}>5^{20}\)
còn câu b thì mk chưa tính ra
4^32=16^16
mà 16^16>16^15
suy ra 4^32>16^15
GTNN của A =2 khi x =3
\(\left(2x+3\right)^{1998}+\left(3y-5\right)^{2000}\le0\)
\(\left\{{}\begin{matrix}\left(2x+3\right)^{1998}\ge0\\\left(3y-5\right)^{2000}\ge0\end{matrix}\right.\)
\(\Rightarrow\left(2x+3\right)^{1998}+\left(3y-5\right)^{2000}\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x+3\right)^{1998}+\left(3y-5\right)^{2000}\ge0\\\left(2x+3\right)^{1998}+\left(3y-5\right)^{2000}\le0\end{matrix}\right.\)
\(\Rightarrow\left(2x+3\right)^{1998}+\left(3y-5\right)^{2000}=0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left(2x+3\right)^{1998}=0\Rightarrow2x+3=0\Rightarrow2x=-3\Rightarrow x=-\dfrac{3}{2}\\\left(3y-5\right)^{2000}=0\Rightarrow3y-5=0\Rightarrow3y=5\Rightarrow y=\dfrac{5}{3}\end{matrix}\right.\)
2)
\(\left(-16\right)^{11}=-\left[\left(2^4\right)^{11}\right]=-\left(2^{44}\right)\)
\(\left(-32\right)^9=-\left[\left(2^5\right)^9\right]=-\left(2^{45}\right)\)
\(-\left(2^{44}\right)>-\left(2^{45}\right)\Rightarrow\left(-16\right)^{11}>\left(-32\right)^9\)
\(\left(2^2\right)^3=2^8\)
\(2^{2^3}=2^8\)
\(2^8=2^8\Rightarrow\left(2^2\right)^3=2^{2^3}\)
\(2^{3^2}=2^9\)
\(2^{2^3}=2^8\)
\(2^9>2^8\Rightarrow2^{3^2}>2^{2^3}\)
a) \(2^{24}=2^{3.8}=8^8\) \(3^{16}=3^{2.8}=9^8\)
Do \(8^8< 9^8\)=> \(2^{24}< 3^{16}\)
b) \(3^{200}=3^{2.100}=9^{100}\); \(2^{300}=2^{3.100}=8^{100}\)
Do \(9^{100}>8^{100}\)=> \(3^{200}>2^{300}\)
c) \(7^{20}=7^{4.5}=2401^5>71^5\)
Vậy \(7^{20}>71^5\)
d) \(\left(-2\right)^{30}=2^{30}=2^{3.10}=8^{10}\); \(\left(-3\right)^{20}=3^{20}=3^{2.10}=9^{10}\)
Do \(8^{10}< 9^{10}\)nên \(\left(-2\right)^{30}< \left(-3\right)^{20}\)
e) \(\left(-5\right)^9< 0\); \(\left(-2\right)^{18}=2^{18}>0\)
Vậy \(\left(-5\right)^9< \left(-2\right)^{18}\)
a) \(3^{21}\)và \(2^{31}\)
\(3^{21}\)=\(3.3^{20}\)=\(3.9^{10}\)
\(2^{31}=2.2^{30}=2.8^{10}\)
Vì \(3.9^{10}\)>\(2.8^{10}\)\(\Rightarrow3^{21}>2^{31}\)
b)\(2^{300}\)và \(3^{200}\)
\(2^{300}=2^{3.100}=\left(2^3\right)^{100}=8^{100}\)
\(3^{200}=3^{2.100}=\left(3^2\right)^{100}=9^{100}\)
Vì \(8^{100}< 9^{100}\Rightarrow2^{300}< 3^{200}\)
c)\(32^9\)và\(18^{13}\)
\(32^9=2^{5.9}=2^{45}\)
\(18^{13}>16^{13}=2^{4.13}=2^{52}\)
\(\Rightarrow2^{45}< 2^{52}< 18^{13}\)\(\Rightarrow2^{45}< 18^{13}\Rightarrow32^9< 18^{13}\)
a) ta có: 321 = 3.320 = 3.910
231 = 2.230 = 2.810
vì 2.810 < 3.910 => 231 < 321
b) ta có: 2300 = (23)100 = 8100
3200 = (32)100 = 9100
vì 8100 < 9100 => 2300 < 3200
c) ta có: 329 = (25)9 = 245
1813 > 1613 = (24)13 = 252
ta thấy 245 < 252 < 1813
Nên 329 < 1813
câu c là (2\(^2\))\(^3\)và 2 mũ 2 mũ 3 nha
giúp mk nhanh 1 chút mk cần rất rất gấp