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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1=1^2 ;125=5^3 ;-125=-5^3; 27=27^1; -27= -27^1
2: 5^2 25^1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(125=5^3\)
\(-125=\left(-5\right)^3\)
\(27=3^3\)
\(-27=\left(-3\right)^3\)
^...^
^_^
Bài 11: Viết các số sau dưới dạng lũy thừa với số mũ khác 1:
125 = 53
(-125) = (-5)3
27 = 33
(-27) = (-3)3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(125=5^3\)
\(-125=\left(-5^3\right)\)
\(27=3^3\)
\(-27=\left(-3^3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
0,25 = (0,5)2
\(\dfrac{1}{49}\) = (\(\dfrac{1}{7}\))2
- \(\dfrac{27}{125}\) = (- \(\dfrac{3}{5}\))3
\(\dfrac{81}{16}\) = ( \(\dfrac{9}{4}\))2
\(\dfrac{169}{196}\) = (\(\dfrac{13}{14}\))2
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
a) \(\left(x-\frac{1}{2}\right)^2=0\)
\(\Rightarrow x-\frac{1}{2}=0\)
\(\Rightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
b) \(\left(x-2\right)^2=1\)
\(\Rightarrow x-2=\pm1\)
+) \(x-2=1\Rightarrow x=3\)
+) \(x-2=-1\Rightarrow x=1\)
Vậy \(x=3\) hoặc \(x=1\)
c) \(\left(2x-1\right)^3=-8\)
\(\Rightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Rightarrow2x-1=-2\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=\frac{-1}{2}\)
Vạy \(x=\frac{-1}{2}\)
d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)
\(\Rightarrow x+\frac{1}{2}=\frac{1}{4}\)
\(\Rightarrow x=\frac{-1}{4}\)
Vậy \(x=\frac{-1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(27\cdot5^3\cdot3^3\cdot32^{-1}:125=3^3\cdot5^3\cdot3^3\cdot\dfrac{1}{32}:5^3=\dfrac{3^6}{32}=\dfrac{3^6}{2^5}\)
125 = 1252-1 ; - 125 = - 1251009-1008 ; 27 = 2710000000000-9999999999 ; - 27 = - 27\(\infty+1-\infty\) :)