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\(\left(-32\right)^9=-\left(2^5\right)^9=-\left(2^{45}\right)\)
\(\left(-16\right)^{13}=-\left(2^4\right)^{13}=-\left(2^{52}\right)\)
vì -2^45>-2^52hay -16^13>-32^9
\(\left(\frac{1}{16}\right)^{10}\) và \(\left(\frac{1}{2}\right)^{50}\)
Ta có: \(\left(\frac{1}{2}\right)^{50}=\left[\left(\frac{1}{2}\right)^5\right]^{10}=\left(\frac{1}{32}\right)^{10}\)
Do \(\frac{1}{6}>\frac{1}{32}\Rightarrow\left(\frac{1}{6}\right)^{10}>\left(\frac{1}{32}\right)^{10}\)
Vậy \(\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
a) \(10^{20}\) và \(9^{10}\)
Vì 10 > 9 ; 20 > 10
nên \(10^{20}>9^{10}\)
Vậy \(10^{20}>9^{10}\)
b) \(\left(-5\right)^{30}\) và \(\left(-3\right)^{50}\)
Ta có: \(\left(-5\right)^{30}=5^{30}=\left(5^3\right)^{10}=125^{10}\)
\(\left(-3\right)^{50}=3^{50}=\left(3^5\right)^{10}=243^{10}\)
Vì 243 > 125 nên \(125^{10}< 243^{10}\)
Vậy \(\left(-5\right)^{30}< \left(-3\right)^{50}\)
c) \(64^8\) và \(16^{12}\)
Ta có: \(64^8=\left(4^3\right)^8=4^{24}\)
\(16^{12}=\left(4^2\right)^{12}=4^{24}\)
Vậy \(64^8=16^{12}\left(=4^{24}\right)\)
d) \(\left(\frac{1}{6}\right)^{10}\) và \(\left(\frac{1}{2}\right)^{50}\)
Ta có: \(\left(\frac{1}{6}\right)^{10}=\left[\left(\frac{1}{2}\right)^4\right]^{10}=\left(\frac{1}{2}\right)^{40}\)
Vì 40 < 50 nên \(\left(\frac{1}{2}\right)^{40}< \left(\frac{1}{2}\right)^{50}\)
Vậy \(\left(\frac{1}{16}\right)^{10}< \left(\frac{1}{2}\right)^{50}\)
bài 2
làm câu B;C nha
B)
\(27^3=\left(3^3\right)^3=3^9\)
\(9^5=\left(3^2\right)^5=3^{10}\)
vì \(10>9\)
\(=>9^5>27^3\)
C)
\(\left(\frac{1}{8}\right)^6=\left(\frac{1}{2^3}\right)^6=\frac{1^6}{2^{18}}=\frac{1}{2^{18}}\)
\(\left(\frac{1}{32}\right)^4=\left(\frac{1}{2^5}\right)^4=\frac{1^4}{2^{20}}=\frac{1}{2^{20}}\)
vì \(2^{18}< 2^{20}\)
\(=>\frac{1}{2^{18}}>\frac{1}{2^{20}}\)
\(=>\left(\frac{1}{8}\right)^6>\left(\frac{1}{32}\right)^4\)
\(\text{A.}\frac{32^3.9^5}{8^3.6^6}=\frac{\left(2^5\right)^3.\left(3^2\right)^5}{\left(2^3\right)^3.\left(2.3\right)^6}=\frac{2^{15}.3^{10}}{2^9.2^6.3^6}=\frac{3^{10}}{3^6}=3^4=81\)
\(\text{B.}\frac{\left(5^5-5^4\right)^3}{50^6}=\frac{2500^3}{50^6}=\frac{\left(50^2\right)^3}{50^6}=\frac{50^6}{50^6}=1\)
Bài 2:
\(\text{A.Ta có:}\)
\(5^6=\left(5^3\right)^2=125^2\)
\(\left(-2\right)^{14}=2^{14}=\left(2^7\right)^2=128^2\)
Vì \(125< 128\)
\(\Rightarrow125^2< 128^2\)
\(\Rightarrow5^6< \left(-2\right)^{14}\)
\(\text{B.Ta có:}\)
\(9^5=\left(3^2\right)^5=3^{10}\)
\(27^3=\left(3^3\right)^3=3^9\)
Vì \(9< 10\)
\(\Rightarrow3^9< 3^{10}\)
\(\Rightarrow27^3< 9^5\)
\(\text{C.Ta có:}\)
\(\left(\frac{1}{8}\right)^6=\left[\left(\frac{1}{2}\right)^3\right]^6=\left(\frac{1}{2}\right)^{18}\)
\(\left(\frac{1}{32}\right)^4=\left[\left(\frac{1}{2}\right)^5\right]^4=\left(\frac{1}{2}\right)^{20}\)
Vì \(18< 20\)
\(\Rightarrow\left(\frac{1}{2}\right)^{18}< \left(\frac{1}{2}\right)^{20}\)
\(\Rightarrow\left(\frac{1}{8}\right)^6< \left(\frac{1}{32}\right)^4\)
Toán 6 ?
Ta có :
\(\left(-\frac{1}{16}\right)^{100}=\left(\frac{1}{16}\right)^{100}=\frac{1}{16^{100}}\)
\(\left(-\frac{1}{2}\right)^{500}=\left(\frac{1}{2}\right)^{500}=\frac{1}{2^{500}}=\frac{1}{\left(2^4\right)^{125}}=\frac{1}{16^{125}}\)
Do \(\frac{1}{16^{100}}>\frac{1}{16^{125}}\left(16^{100}< 16^{125}\right)\)
\(\Rightarrow\left(-\frac{1}{16}\right)^{100}>\left(-\frac{1}{.2}\right)^{500}\)
Vậy ...
a) \(\left(-\frac{1}{2}\right)^{500}=\left[\left(-\frac{1}{2}^5\right)^{100}\right]=\left(\frac{-1}{32}\right)^{100}\)
Vì \(\left(-\frac{1}{16}\right)^{100}\) > \(\left(\frac{-1}{32}\right)^{100}\) nên \(\left(-\frac{1}{16}\right)^{100}>\left(-\frac{1}{2}\right)^{500}\)
b) Câu này mk ko bt
Bạn thông cảm
1.\(45^{10}.5^{30}=45^{10}.125^{10}=\left(45.125\right)^{10}=5625^{10}\)
2.a. \(\left(2x-1\right)^3=-8\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow2x-1=-2\Leftrightarrow x=-\frac{1}{2}\)
b.\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}}\)
c. \(\left(2x+3\right)^2=\frac{9}{121}\Leftrightarrow\orbr{\begin{cases}2x+3=\frac{3}{11}\\2x+3=-\frac{3}{11}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{15}{11}\\x=-\frac{18}{11}\end{cases}}\)
d.\(\left(3x-1\right)^3=-\frac{8}{27}=\left(-\frac{2}{3}\right)^3\)
\(\Leftrightarrow3x-1=-\frac{2}{3}\Leftrightarrow x=\frac{1}{9}\)
4.
a.\(99^{20}=\left(99^2\right)^{10}=9801^{10}\)
Do \(9801^{10}< 9999^{10}\Rightarrow99^{20}< 9999^{10}\)
b.\(3^{4000}=\left(3^2\right)^{2000}=9^{2000}\)
\(\Rightarrow3^{4000}=9^{2000}\)
c.\(2^{332}=\left(2^3\right)^{110}.2^2=8^{110}.4\)
\(3^{223}=\left(3^2\right)^{110}.3^3=\left(3^2\right)^{110}.9=9^{110}.9\)
Ta thấy \(4.8^{110}< 9.9^{110}\)
Vậy \(2^{332}< 3^{223}\)
Bài 1: a) (2x+1)2 = 25
(2x+1)2 = 52
=> 2x + 1 = 5 hoặc 2x+1 = -5
=> x=2 hoặc x=-3
b) 2x+2 - 2x = 96
<=> 2x . 22 - 2x = 96
<=> 2x(4-1) =96
<=>2x = 96 :3 = 32 = 25
<=> x = 5
c) (x-1)3 = 125
<=> (x-1)3 = 53
<=> x-1=5
<=>x= 5 +1 = 6
Bài 2 :2
a) 3200 và 2300
Ta có :
3200 = ( 32 )100 = 9100
2300 = ( 23 )100 = 8100
Vì 9100 > 8100 Nên 3200 > 2300
b) 1255 và 257
Ta có :
1255 = ( 53 )5 = 515
257 = ( 52 )7 = 514
Vì 515 > 514 ( 15 > 14 )
Nên 1255 > 257
Tương tự ....
Bài 2 :
a) 3200 và 2300
Ta có :
3200 = ( 32 )100 = 9100
2300 = ( 23 )100 = 8100
Vì 9100 > 8100 nên 3200 > 2300
bt mỗi câu này thôi
a) \(\left(-32\right)^9=\left[\left(-2\right)^5\right]^9=\left(-2\right)^{45}\)
\(\left(-16\right)^{13}=\left[-2^4\right]^{13}=-2^{52}\)
Vì \(-2^{45}>-2^{52}\Rightarrow-32^9>-16^{13}\)
ý b,c làm tương tự a) bạn nhé ! Có gì thắc mắc thì để lại bình luận...
b) \(\left(-5\right)^{30}< \left(-3\right)^{50}\)
c) \(\left(-\dfrac{1}{8}\right)^{1000}< \left(-\dfrac{1}{2}\right)^{500}\)
Thanks bạn nhé!