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Ta có: 2515= (52)15= 530
810 . 330= (23)10. 330= 230. 330= (2. 3)30= 630
Ta thấy: 5< 6
=> 530< 630
=> 2515< 810. 330
Vậy 2515< 810. 330
Dấu "." là dấu nhân nha( vì mik ko viết được dấu nhân)!!!!
Chúc bạn học tốt!!!!!
ta có :\(25^{15}=5^{30}\)(1)
\(8^{10}\times3^{30}=2^{30}\times3^{30}=6^{30}\)(2)
từ (1) vả (2) \(\Rightarrow5^{30}< 6^{30}\Rightarrowđpcm\)
Ta có : 333^444=(3.111)^444=3^444.111^444
444^333=(4.111)^333=4^333.111^333
Ta lại có : 3^444=(3^4)^111=81^111
4^333=(4^3)^111=64^111
vì 3^444>4^333
mặt khác 111^333<111^444
suy ra 4^333.111^333<3^444.111^444
vậy 333^444>444^333
\(\frac{1}{2}.2^n+4.2^n=9.2^5\Rightarrow2^n\left(\frac{1}{2}+4\right)=288\Rightarrow2^n.\frac{9}{2}=288\Rightarrow2^{n-2}.9=288\Rightarrow2^{n-2}=32\)(dấu "=>" số 3 bn sửa thành 2n-1.9=288=>2n-1=32 nha)
=>2n-1=25=>n-1=5=>n=5+1=6
vậy......
~~~~~~~~~~~~~~~
a: \(\Leftrightarrow4^x\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)=4^8\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)\)
=>4^x=4^8
=>x=8
b: \(\Leftrightarrow2^x\cdot\dfrac{1}{2}+2^x\cdot2=2^{10}\left(2^2+1\right)\)
=>2^x=2^11
=>x=11
c: =>1/6*6^x+6^x*36=6^15(1+6^3)
=>6^x=6*6^15
=>x=16
d: \(\Leftrightarrow8^x\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)=8^9\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)\)
=>x=9
a, \(4^3.5^3=\left(4.5\right)^3=20^3=8000\)
b, \(6^3.5^3=\left(6.5\right)^3=30^3=27000\)
c, \(8^2.5^2=\left(8.5\right)^2=40^2=1600\)
d, \(125^3.8^3=\left(125.8\right)^3=1000^3\)
e, \(5^2.6^2.3^2=\left(5.6.3\right)^2=90^2\)
a)\(VT=\left(-\dfrac{1}{8}\right)^{100}=\dfrac{1}{8^{100}}=\dfrac{1}{\left(2^3\right)^{100}}=\dfrac{1}{2^{300}}\)
\(VP=\left(-\dfrac{1}{4}\right)^{200}=\dfrac{1}{\left(2^2\right)^{200}}=\dfrac{1}{2^{400}}\)
\(\Rightarrow VT>VP\)
b) \(VT=4^{100}=\left(2^2\right)^{100}=2^{200}< 2^{202}=VP\)
c) \(VT=5^{2000}=\left(5^2\right)^{1000}=25^{1000}>10^{1000}=VP\)
d) \(VT=31^5< 32^5=\left(2^5\right)^5=2^{25}\)
\(VP=17^7>16^7=\left(2^4\right)^7=2^{28}\)
\(VP>VT\)
\(1.a)\) Ta có: \(\left\{{}\begin{matrix}64^8=\left(8^2\right)^8=8^{16}\\16^{12}=8^{12}.2^{12}=8^{12}.\left(2^3\right)^4=8^{12}.8^4=8^{16}\end{matrix}\right.\)
Có: \(8^{16}=8^{16}\Rightarrow64^8=16^{12}\)
Vậy...
\(b)\) Ta có: \(\left\{{}\begin{matrix}\left(-5\right)^{30}=\left[\left(-5\right)^3\right]^{10}=\left(-125\right)^{10}\\\left(-3\right)^{50}=\left[\left(-3\right)^5\right]^{10}=\left(-243\right)^{10}\end{matrix}\right.\)
Có: \(\left(-125\right)^{10}< \left(-243\right)^{10}\Rightarrow\left(-5\right)^{30}< \left(-3\right)^{50}\)
Vậy...
\(c)\) Ta có: \(\left\{{}\begin{matrix}2^{27}=\left(2^3\right)^9=8^9\\3^{18}=\left(3^2\right)^9=9^9\end{matrix}\right.\)
Có: \(8^9< 9^9\Rightarrow2^{27}< 3^{18}\)
Vậy...
\(d)\) Ta có: \(\left\{{}\begin{matrix}\left(\dfrac{1}{25}\right)^{10}=\left[\left(\dfrac{1}{5}\right)^2\right]^{10}=\left(\dfrac{1}{5}\right)^{20}\\\left(\dfrac{1}{125}\right)^8=\left[\left(\dfrac{1}{5}\right)^3\right]^8=\left(\dfrac{1}{5}\right)^{24}\end{matrix}\right.\)
Có: \(\left(\dfrac{1}{5}\right)^{20}< \left(\dfrac{1}{5}\right)^{24}\Rightarrow\left(\dfrac{1}{24}\right)^{10}< \left(\dfrac{1}{125}\right)^8\)
Vậy...
\(e)\)Có: \(32^9=\left(2^5\right)^9=2^{45}< 2^{52}=\left(2^4\right)^{13}=16^{13}< 18^{13}\)
\(\Rightarrow32^9< 18^{13}\)
Vậy...
Bài 1: So Sánh
a) Ta có: \(2^{100}=2^{10^{10}}=1024^{10}\)
\(10^{30}=10^{3\cdot10}=1000^{10}\)
mà \(1024^{10}>1000^{10}\)
nên \(2^{100}>10^3\)
b) Ta có: \(5\cdot8^{25}=5\cdot2^{75}\)
\(128^{11}=2^{77}=4\cdot2^{75}\)
mà \(5\cdot2^{75}>4\cdot2^{75}\)
nên \(5\cdot8^{25}>128^{11}\)
c) Ta có: \(8\cdot27^6=8\cdot3^{18}\)
\(9^{10}=3^{20}=9\cdot3^{18}\)
mà \(8\cdot3^{18}< 9\cdot3^{18}\)
nên \(8\cdot27^6< 9^{10}\)
d) Ta có: \(2^{100}=2^{69}\cdot2^{31}\)
\(=2^{31}\cdot2^{63}\cdot2^6\)
\(=2^{31}\cdot\left(2^9\right)^7\cdot\left(2^2\right)^3\)
\(=2^{31}\cdot512^7\cdot4^3\)
Ta có: \(10^{31}=2^{31}\cdot5^{31}\)
\(=2^{31}\cdot5^{28}\cdot5^3\)
\(=2^{31}\cdot\left(5^4\right)^7\cdot5^3\)
\(=2^{31}\cdot625^7\cdot5^3\)
Ta có: \(512^7< 625^7\)
\(4^3< 5^3\)
Do đó: \(512^7\cdot4^3< 625^7\cdot5^3\)
\(\Leftrightarrow2^{31}\cdot512^7\cdot4^3< 2^{31}\cdot625^7\cdot5^3\)
hay \(2^{100}< 10^{31}\)