Bài 6: Lũy thừa của một số hữu tỉ (tiếp theo...)

a) \(2^x=8\)

⇔ \(2^x=2^3\)

⇒ \(x=3\)

b) \(3^x=27\)

⇔ \(3^x=3^3\)

⇒ \(x=3\)

c) \(\left(-\dfrac{1}{2}\right)x=\left(-\dfrac{1}{2}\right)^4\)

⇔ \(x=\left(-\dfrac{1}{2}\right)^4\div\left(-\dfrac{1}{2}\right)\)

⇔ \(x=\left(-\dfrac{1}{2}\right)^3\)

d) \(x\div\left(-\dfrac{3}{4}\right)=\left(-\dfrac{3}{4}\right)^2\)

⇔ \(x=\left(-\dfrac{3}{4}\right)^2\cdot\left(-\dfrac{3}{4}\right)\)

⇔ \(x=\left(-\dfrac{3}{4}\right)^3=-\dfrac{27}{64}\)

d) \(\left(x+1\right)^3=-125\)

⇔ \(\left(x+1\right)^3=\left(-5\right)^3\)

⇔ \(x+1=-5\)

⇔ \(x=-5-1=-6\)

2:

a: (x-1,2)^2=4

=>x-1,2=2 hoặc x-1,2=-2

=>x=3,2(loại) hoặc x=-0,8(loại)

b: (x-1,5)^2=9

=>x-1,5=3 hoặc x-1,5=-3

=>x=-1,5(loại) hoặc x=4,5(loại)

c: (x-2)^3=64

=>(x-2)^3=4^3

=>x-2=4

=>x=6(nhận)

\(4^{15}.9^{15}=\left(4.9\right)^{15}=36^{15}=\left(6^2\right)^{15}=6^{30}\\ 18^{16}.2^{16}=\left(18.2\right)^{16}=36^{16}=\left(6^2\right)^{16}=6^{32}\\ Vậy:2^n.3^n=6^n\\ Vậy:6^{30}< 6^n< 6^{32}\\ Vậy:n=31\)

\(\left(x+0,8\right)^2=0,25\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+0,8\right)^2=\left(0,5\right)^2\\\left(x+0,8\right)^2=\left(-0,5\right)^2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+0,8=0,5\\x+0,8=-0,5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0,5-0,8=-0,3\\x=-0,5-0,8=-1,3\end{matrix}\right.\)

a) \(\left(2x-3\right)^4=5^4\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=5\\2x-3=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=8\\2x=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{2}=4\\x=-\dfrac{2}{2}=-1\end{matrix}\right.\)

b) \(\left(2x-3\right)^3=-64\)

\(\Rightarrow\left(2x-3\right)^3=\left(-4\right)^3\)

\(\Rightarrow2x-3=-4\)

\(\Rightarrow2x=-1\)

\(\Rightarrow x=-\dfrac{1}{2}\)

a/

\(\left(2x-3\right)^4=5^4\\ \Leftrightarrow\left|2x-3\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=5\\2x-3=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=8\\2x=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)

Vậy...

b/ 2x-3=-64

<=>2x = -61

<=> x = -61/2

\(\left(2x+1\right)^3=-0,001\)

\(\Rightarrow\left(2x+1\right)^3=\left(-0,1\right)^3\)

\(\Rightarrow2x+1=-0,1\)

\(\Rightarrow2x=-0,1-1\)

\(\Rightarrow2x=-1,1\)

\(\Rightarrow x=\dfrac{-1,1}{2}\)

\(\Rightarrow x=-0,55\)

\(\dfrac{5^{102}\cdot9^{1009}}{3^{2018}\cdot25^{50}}\)

\(=\dfrac{5^{102}\cdot\left(3^2\right)^{1009}}{3^{2018}\cdot\left(5^2\right)^{50}}\)

\(=\dfrac{5^{102}\cdot3^{2018}}{3^{2018}\cdot5^{100}}\)

\(=\dfrac{5^2\cdot1}{1\cdot1}\)

\(=25\)

25^50=(5^2)^50=5^100

`25^50 =(5^2 )^50 =5^(2.50)=5^100`

25^50=(5^2)^50=5^100

Ta có:

\(5^{102}=5^{2.51}=\left(5^2\right)^{51}=25^{51}\)

\(\Rightarrow5^{102}=25^{51}\)

Ta có 5¹⁰² = 5^2.51 = (5²)⁵¹ = 25⁵¹

⇒ 5¹⁰² = 25⁵¹

13:

x^3/2=4

=>x^3=8

=>x=2

14: x^3/-3=9

=>x^3=-27

=>x=-3

15: x^5:x^3=1/16

=>x^2=1/16

=>x=1/4 hoặc x=-1/4

16: x^3=x

=>x(x^2-1)=0

=>x(x-1)(x+1)=0

=>x=0;x=1;x=-1

17: =>x^3=1/27

=>x=1/3

18: =>x^2=9^2=81

=>x=9 hoặc x=-9

19: =>x+2=-2 hoặc x+2=2

=>x=0 hoặc x=-4

20: =>2x-3=5 hoặc 2x-3=-5

=>2x=-2 hoặc 2x=8

=>x=4 hoặc x=-1

21: =>x+4=7 hoặc x+4=-7

=>x=-11 hoặcx=3

\(=0.04\cdot5-\dfrac{2^6\cdot3^{10}}{3^9\cdot2^6}=0.2-3=-2.8\)

Yc của đề là gì?