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14 tháng 4 2021
4^2x-6=1 =>4^2x-6=4^0 =>2x-6=0 =>2x=6 =>x=6÷2 =>x=3
6 tháng 8 2017

1. 2x=16\(\Rightarrow\)X=4

2. 22x-1=27

\(\Rightarrow\)27=22.4-1

Vậy x =4

x=4 nha chị

6 tháng 10 2020

(2x - 1)6 = (2x - 1)4

=> (2x - 1)6 - (2x - 1)4 = 0

=> (2x - 1)4.[(2x - 1)2 - 1] = 0

=> \(\orbr{\begin{cases}\left(2x-1\right)^4=0\\\left(2x-1\right)^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x-1=0\\\left(2x-1\right)^2=1\end{cases}}\Rightarrow\orbr{\begin{cases}2x-1=0\\2x-1=\pm1\end{cases}}\)

Khi 2x - 1 = 0 => x = 1/2

Khi 2x - 1 = -1 => x = 0

Khi 2x - 1 = 1 => x = 1

Vậy \(x\in\left\{\frac{1}{2};0;1\right\}\)là giá trị cần tìm

6 tháng 10 2020

( 2x - 1 )6 = ( 2x - 1 )4

<=> ( 2x - 1 )6 - ( 2x - 1 )4 = 0

<=> ( 2x - 1 )4[ ( 2x - 1 )2 - 1 ] = 0

<=> \(\orbr{\begin{cases}\left(2x-1\right)^4=0\\\left(2x-1\right)^2-1=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=\frac{1}{2}\\x=1\\x=0\end{cases}}\)( thay = dấu hoặc hộ nhé )

23 tháng 8 2023

a) \(4^n=4096\Rightarrow4^n=4^6\Rightarrow n=6\)

b) \(5^n=15625\Rightarrow5^n=5^6\Rightarrow n=6\)

c) \(6^{n+3}=216\Rightarrow6^{n+3}=6^3\Rightarrow n+3=3\Rightarrow n=0\)

d) \(x^2=x^3\Rightarrow x^3-x^2=0\Rightarrow x^2\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

e) \(3^{x-1}=27\Rightarrow3^{x-1}=3^3\Rightarrow x-1=3\Rightarrow x=4\)

f) \(3^{x+1}=9\Rightarrow3^{x+1}=3^2\Rightarrow x+1=2\Rightarrow x=1\)

g) \(6^{x+1}=36\Rightarrow6^{x+1}=6^2\Rightarrow x+1=2\Rightarrow x=1\)

h) \(3^{2x+1}=27\Rightarrow3^{2x+1}=3^3\Rightarrow2x+1=3\Rightarrow2x=2\Rightarrow x=1\)

i) \(x^{50}=x\Rightarrow x^{50}-x=0\Rightarrow x\left(x^{49}-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^{49}-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x^{49}=1=1^{49}\end{matrix}\right.\)  \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

23 tháng 8 2023

4n  =  4096 

4n = 212

n = 12

5n = 15625 

5n = 56

n   = 6

6n+3 = 216

6n+3 = 23.33

6n+3 = 63

n + 3 = 3

 

 

22 tháng 10 2019

1. \(6x^3-8=40\\ 6x^3=48\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2

2. \(4x^5+15=47\\ 4x^5=32\\ x^5=8\\ \Rightarrow x\in\varnothing\left(\text{vì }x\in N\right)\)Vậy x ∈ ∅

3. \(2x^3-4=12\\ 2x^3=16\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2

4. \(5x^3-5=0\\ 5x^3=5\\ x^3=1\\ \Rightarrow x=1\)Vậy x = 1

5. \(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)Vậy \(x\in\left\{5;6\right\}\)

6. \(\left(3x-2\right)^{20}=\left(3x-1\right)^{20}\\ \Rightarrow3x-2=3x-1\\ 3x-3x=2-1\\ 0=1\left(\text{vô lí}\right)\)Vậy x ∈ ∅

7. \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\\ \left(3x-1\right)^{10}=\left[\left(3x-1\right)^2\right]^{10}\\ \Rightarrow\left(3x-1\right)^2=3x-1\\ \left(3x-1\right)^2-\left(3x-1\right)=0\\ \left(3x-1\right)\left[\left(3x-1\right)-1\right]=0\\ \left(3x-1\right)\left(3x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-1=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\left(\text{loại vì }x\in N\right)\\x=\frac{2}{3}\left(\text{loại vì }x\in N\right)\end{matrix}\right.\)Vậy x ∈ ∅

8. \(\left(2x-1\right)^{50}=2x-1\\ \left(2x-1\right)^{50}-\left(2x-1\right)=0\\ \left(2x-1\right)\left[\left(2x-1\right)^{49}-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^{49}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=1\\2x-1=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(\text{loại vì }x\in N\right)\\x=1\left(t/m\right)\end{matrix}\right.\)Vậy x = 1

9. \(\left(\frac{x}{3}-5\right)^{2000}=\left(\frac{x}{3}-5\right)^{2008}\\ \left(\frac{x}{3}-5\right)^{2008}-\left(\frac{x}{3}-5\right)^{2000}=0\\ \left(\frac{x}{3}-5\right)^{2000}\left[\left(\frac{x}{3}-5\right)^8-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(\frac{x}{3}-5\right)^{2000}=0\\\left(\frac{x}{3}-5\right)^8=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}-5=0\\\frac{x}{3}-5=1\\\frac{x}{3}-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}=5\\\frac{x}{3}=6\\\frac{x}{3}=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\cdot3=15\\x=6\cdot3=18\\x=4\cdot3=12\end{matrix}\right.\)Vậy \(x\in\left\{15;18;12\right\}\)

22 tháng 10 2019

\(1.6x^3-8=40\\ \Leftrightarrow6x^3=48\\ \Leftrightarrow x^3=8\Leftrightarrow x^3=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{2;-2\right\}\)

\(2.4x^3+15=47\) (T nghĩ đề là mũ 3)

\(\Leftrightarrow4x^3=32\Leftrightarrow x^3=8=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{2;-2\right\}\)

Câu 3, 4 tương tự nhé.

2 tháng 8 2023

chịu

19 tháng 12 2023

Chịu 

30 tháng 7 2023

1) \(2^8.4.16.32.64.2=2^8.2^2.2^4.2^5.2^6.2=2^{8+2+4+5+6+1}=2^{26}\)

2) \(7.3^x-11=52\Rightarrow7.3^x=52+11=63\)

\(\Rightarrow3^x=63:7=9\Rightarrow3^x=3^2\Rightarrow x=2\)

\(4^{10\left(x+5\right)}=4^{12}\Rightarrow10\left(x+5\right)=12\)

\(\Rightarrow10x+50=12\Rightarrow10x=12-50=-38\)

\(\Rightarrow x=-38:10=-\dfrac{38}{10}=-\dfrac{19}{5}\)

\(7x-2x=6^{17}:6^{15}+2^8:2^6\)

\(\Rightarrow5x=6^2+2^2\Rightarrow5x=36+4=40\)

\(\Rightarrow x=40:5=8\)

\(3^{2x-1}=27\Rightarrow3^{2x-1}=3^3\Rightarrow2x-1=3\)

\(\Rightarrow2x=4\Rightarrow x=4:2=2\)

30 tháng 7 2023

1) 28. 4 . 16 . 32 . 64

= 28. 22. 24. 25. 26

= 225

26 tháng 9 2020

a,\(\left(2x+1\right)^3=125\)

\(\left(2x+1\right)^3=5^3\)

\(2x+1=5\)

\(2x=4\)

\(x=2\)

a: =>2x^3=58-4=54

=>x^3=27

=>x=3

b; =>(5-x)^5=2^5

=>5-x=2

=>x=3

c: =>(5x-6)^3=4^3

=>5x-6=4

=>5x=10

=>x=2

d: (3x)^3=(2x+1)^3

=>3x=2x+1

=>x=1

12 tháng 8 2023

1=>2x3=54
=>x3=27  =>x=3
2=>(5-x)5=25
=>5-x=2
=>x=3
3=>(5x-6)3=43
=>5x-6=4
=>5x=10=>x=2
4=>3x=2x+1
=>x=1

25 tháng 12 2021

gúp minh với ạ 

30 tháng 11 2016

Ta có:\(\left(x-4\right)\left(x-1\right)^2=0\)

\(\Leftrightarrow x-4=0\) hoặc \(\left(x-1\right)^2=0\)

\(\Leftrightarrow x=4\) hoặc \(x-1=0\)

\(\Leftrightarrow x=4\) hoặc x=1

\(\left(2x-6\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow2x-6=0\) hoặc \(x^2+1=0\)(vô lí)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)

Xét VT ta có:1+2+3+4+...+x=\(\frac{x\left(x+1\right)}{2}\)

Mà VP=78

\(\Rightarrow\frac{x\left(x+1\right)}{2}=78\)

\(\Rightarrow x\left(x+1\right)=156\)

\(x\left(x+1\right)=12\cdot13\)

\(\Rightarrow x=12\)

 

vt la gi va vp la gi