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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: \(\left(-\frac{1}{16}\right)=-\left(\frac{1}{2}\right)^4.\)
\(\left(-\frac{1}{2}\right)=-\left(\frac{1}{2}\right)^1< -\left(\frac{1}{2}\right)^4\)
b) \(9^5=3^{2\cdot5}=3^{10}\)
\(27^3=3^{3\cdot3}=3^9\)
=> tự kết luận
c) \(\left(\frac{1}{8}\right)^6=\left(\frac{1}{2}^3\right)^6=\left(\frac{1}{2}\right)^{18}\)
\(\left(\frac{1}{32}\right)^4=\left(\frac{1}{2}^5\right)^4=\left(\frac{1}{2}\right)^{20}\)
=> tự kết luận
b) Ta có: \(9^5=\left(3^2\right)^5=3^{10}\)
\(27^3=\left(3^3\right)^3=3^9\)
Vì 10 > 9 => 310 > 39
Vậy 95 > 273
1. So sánh :
b) 9^5 và 27^3
9^5 = ( 3^2 )^5 = 3^10
27^3 = ( 3^3 )^3 = 3^9
Vì 3^10 > 3^9 => 9^5 > 27^3
Vậy 9^5 > 27^3
c) \(\left(\frac{1}{8}\right)^6\)và \(\left(\frac{1}{32}\right)^4\)
\(\left(\frac{1}{8}\right)^6=\left(\frac{1}{2}\right)^{3.6}=\left(\frac{1}{2}\right)^{18}\)
\(\left(\frac{1}{32}\right)^4=\left(\frac{1}{2}\right)^{5.4}=\left(\frac{1}{2}\right)^{20}\)
Vì ( 1/2)^18 < (1/2)^20 => (1/8)^6 < (1/32)^4
Vậy (1/8)^6 < (1/32)^4
Bài 1:
\(a,A=\frac{-25}{28}.0,21=\frac{-25}{28}.\frac{21}{100}=\frac{-25.21}{28.100}=\frac{-1.25.3.7}{4.7.25.4}=\frac{-1.3}{4.4}=\frac{-3}{16}\)
\(b,B=\left(\frac{13}{24}-\frac{29}{30}\right):\left(-10,2\right)=\left(\frac{65}{120}-\frac{116}{120}\right):\frac{-51}{5}=\frac{-51}{120}.\frac{5}{-51}=\frac{-51.5}{120.\left(-51\right)}=\frac{-51.5}{5.24.\left(-51\right)}=\frac{1}{24}\)
a) \(\left|2x+\frac{3}{4}\right|=\frac{1}{2}\)
\(\orbr{\begin{cases}2x+\frac{3}{4}=\frac{1}{2}\\2x+\frac{3}{4}=\frac{-1}{2}\end{cases}}\) => \(\orbr{\begin{cases}2x=\frac{1}{2}-\frac{3}{4}\\2x=\frac{-1}{2}-\frac{3}{4}\end{cases}}\) => \(\orbr{\begin{cases}2x=\frac{-1}{4}\\2x=\frac{-5}{4}\end{cases}}\) => \(\orbr{\begin{cases}x=\frac{-1}{8}\\x=\frac{-5}{8}\end{cases}}\)
Vậy \(x=\left\{\frac{-1}{8},\frac{-5}{8}\right\}\)
b) \(\frac{3x}{2,7}=\frac{\frac{1}{4}}{2\frac{1}{4}}\)= \(\frac{3x}{2,7}=\frac{\frac{1}{4}}{\frac{9}{4}}\)
=> \(3x.\frac{9}{4}=2,7.\frac{1}{4}\)=> \(\frac{27x}{4}=\frac{27}{40}\)
\(27x.40=27.4\)
\(1080.x=108\)
\(x=\frac{1}{10}\)
Vậy \(x=\frac{1}{10}\)
c) \(\left|x-1\right|+4=6\)
\(\left|x-1\right|=6-4\)
\(\left|x-1\right|=2\)
\(\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\)=> \(\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
Vậy \(x=\left[3,-1\right]\)
d) \(\frac{x}{3}=\frac{y}{5}=>\frac{y}{5}=\frac{x}{3}=>\frac{y-x}{5-3}=\frac{24}{2}=12\)
e) \(\left(x^2-3\right)^2=16\)
\(\left(x^2-3\right)^2=4^2\)\(=>x^2-3=4\)
\(x^2=7=>x=\sqrt{7}\)
Vậy \(x=\sqrt{7}\)
f) \(\frac{3}{4}+\frac{2}{5}x=\frac{29}{60}\)
\(\frac{2}{5}x=\frac{29}{60}-\frac{3}{4}\)
\(\frac{2}{5}x=-\frac{4}{15}\)
\(x=-\frac{4}{15}:\frac{2}{5}=-\frac{4}{15}.\frac{5}{2}=-\frac{2}{3}\)
Vậy \(x=-\frac{2}{3}\)
g) \(\left(-\frac{1}{3}\right)^3.x=\frac{1}{81}\)
\(\left(-\frac{1}{27}\right).x=\frac{1}{81}\)
\(x=\left(-\frac{1}{27}\right):\frac{1}{81}=\left(-\frac{1}{27}\right).81=-3\)
Vậy \(x=-3\)
k)\(\frac{3}{4}-\frac{2}{5}x=\frac{29}{60}\)
\(\frac{2}{5}x=\frac{3}{4}-\frac{29}{60}\)
\(\frac{2}{5}x=\frac{4}{15}\)
\(x=\frac{2}{5}-\frac{4}{15}=>x=\frac{2}{15}\)
Vậy \(x=\frac{2}{15}\)
I) \(\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
\(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}\)
\(\frac{3}{5}x=\frac{5}{14}\)
\(x=\frac{5}{14}:\frac{3}{5}=\frac{5}{14}.\frac{5}{3}=\frac{25}{42}\)
Vậy \(x=\frac{25}{42}\)
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: \(\left(\frac{1}{16}\right)^{10}=\left(\frac{1}{2^4}\right)^{10}=\frac{1}{2^{40}}\)
\(\left(\frac{1}{2}\right)^{50}=\frac{1}{2^{50}}\)
\(\Rightarrow\frac{1}{2^{40}}>\frac{1}{2^{50}}\)
\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
mãi ko thấy ai làm tớ làm giúp nì =)
\(\text{ta có:}\hept{\begin{cases}\frac{2002}{2003}< 1\\\frac{2005}{2004}>1\end{cases}}\Rightarrow\frac{2005}{2004}>\frac{2002}{2003}\Rightarrow-\frac{2005}{2004}< -\frac{2002}{2003}\)
\(\text{ta có: }\hept{\begin{cases}-\frac{1}{10^5}< 0\\\frac{-9}{-10}>0\end{cases}}\Rightarrow\frac{-1}{10^5}< \frac{-9}{-10}\)