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Bài 1 và Bài 2 dễ, bn có thể tự làm được!
Bài 3:
a) ta có: 1020 = (102)10 = 10010
=> 10010>910
=> 1020>910
b) ta có: (-5)30 = 530 =( 53)10 = 12510 ( vì là lũy thừa bậc chẵn)
(-3)50 = 350 = (35)10= 24310
=> 12510 < 24310
=> (-5)30 < (-3)50
c) ta có: 648 = (26)8= 248
1612 = ( 24)12 = 248
=> 648 = 1612
d) ta có: \(\left(\frac{1}{16}\right)^{10}=\left(\frac{1}{2^4}\right)^{10}=\frac{1}{2^{40}}\)
\(\left(\frac{1}{2}\right)^{50}=\frac{1}{2^{50}}\)
\(\Rightarrow\frac{1}{2^{40}}>\frac{1}{2^{50}}\)
\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
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: \(\left(-\frac{1}{16}\right)=-\left(\frac{1}{2}\right)^4.\)
\(\left(-\frac{1}{2}\right)=-\left(\frac{1}{2}\right)^1< -\left(\frac{1}{2}\right)^4\)
Ta có:
\(\left(\frac{1}{16}\right)^{50}=\left[\left(\frac{1}{2}\right)^4\right]^{50}=\left(\frac{1}{2}\right)^{200}=\frac{1^{200}}{2^{200}}=\frac{1}{2^{200}}\)
\(\left(\frac{1}{2}\right)^{60}=\frac{1^{60}}{2^{60}}=\frac{1}{2^{60}}\)
Vì \(2^{200}>2^{60}\Rightarrow\frac{1}{2^{200}}< \frac{1}{2^{60}}\Rightarrow\left(\frac{1}{16}\right)^{50}< \left(\frac{1}{2}\right)^{60}\)
Vì\(\left(\frac{1}{16}\right)^{10}\)= \(\left[\left(\frac{1}{2}\right)^4\right]^{10}\)= \(\left(\frac{1}{2}\right)^{40}\)
Mà 40<50 =>\(\left(\frac{1}{2}\right)^{40}\)< \(\left(\frac{1}{2}\right)^{50}\)hay \(\left(\frac{1}{16}\right)^{10}\)< \(\left(\frac{1}{2}\right)^{50}\)
Vậy \(\left(\frac{1}{16}\right)^{10}\)<\(\left(\frac{1}{2}\right)^{50}\)
Học giỏi!^^ (đúng thì k cho mik nhé,cảm ơn!)
\(\left(\frac{1}{2}\right)^{50}=\left(\left(\frac{1}{2}\right)^5\right)^{10}=\left(\frac{1}{32}\right)^{10}\)
Ta có\(\frac{1}{16}>\frac{1}{32}\)nên\(\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{32}\right)^{10}\)hay\(\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
a: \(=\dfrac{5}{3}\left(-16-\dfrac{2}{7}+28+\dfrac{2}{7}\right)=\dfrac{5}{3}\cdot12=20\)
b: \(=\left(4\cdot\dfrac{3}{4}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17=\dfrac{1}{2}\cdot\dfrac{6}{5}-17=\dfrac{3}{5}-17=-\dfrac{82}{5}\)
c: \(=-\left(\dfrac{1}{3}\right)^{50}\cdot3^{50}-\dfrac{2}{3}\cdot\dfrac{1}{4}=-1-\dfrac{1}{6}=-\dfrac{7}{6}\)
e: \(=5.7\left(-6.5-3.5\right)=-5.7\cdot10=-57\)
a)
Vì 3<5
\(\Rightarrow3^{30}< 5^{30}\)
\(\Rightarrow\left(-3\right)^{30}< \left(-5\right)^{30}\)
b)
Ta có
\(\left(\frac{1}{2}\right)^{50}=\left[\left(\frac{1}{2}\right)^4\right]^{10}.\left(\frac{1}{2}\right)^{10}\)
\(=\left(\frac{1}{16}\right)^{10}.\left(\frac{1}{2}\right)^{10}\)
Ta có
\(\left(\frac{1}{2}\right)^{10}< 1\)
\(\Leftrightarrow\left(\frac{1}{16}\right)^{10}.\left(\frac{1}{2}\right)^{10}< \left(\frac{1}{16}\right)^{10}\)
\(\Leftrightarrow\left(\frac{1}{2}\right)^{50}< \left(\frac{1}{16}\right)^{10}\)
câu a bạn nhầm đề ạ ^^