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25 tháng 10 2023

\(3^{x+2}+3^x=270\\\Rightarrow 3^x\cdot\cdot3^2+3^x=270\\\Rightarrow 3^x\cdot(3^2+1)=270\\\Rightarrow 3^x\cdot(9+1)=270\\\Rightarrow 3^x\cdot10=270\\\Rightarrow 3^x=270:10\\\Rightarrow 3^x=27\\\Rightarrow 3^x=3^3\\\Rightarrow x=3\)

16 tháng 9 2020

\(3^{x+4}+3^{x+2}=270\)  

\(3^{x+2+2}+3^{x+2}=270\) 

\(3^{x+2}\cdot3^2+3^{x+2}\cdot1=270\) 

\(3^{x+2}\left(3^2+1\right)=270\) 

\(3^{x+2}\cdot10=270\) 

\(3^{x+2}=270:10\) 

\(3^{x+2}=27\) 

\(3^{x+2}=3^3\) 

\(x+2=3\) 

\(x=1\)

16 tháng 9 2020

                                                                 Bài giải

\(3^{x+4}+3^{x+2}=270\)

\(3^{x+2}\left(3^2+1\right)=270\)

\(3^{x+2}\cdot10=270\)

\(3^{x+2}=27=3^3\)

\(x+2=3\)

\(x=1\)

AH
Akai Haruma
Giáo viên
4 tháng 2 2023

Đề bị lỗi. Bạn coi lại đề.

7 tháng 8 2024

  Bài 1:

2\(x\) = 4

2\(^x\) = 22

 \(x=2\)

Vậy \(x=2\)

7 tháng 8 2024

Bài 2:

2\(^x\) = 8

2\(^x\) = 23

\(x=3\)

Vậy \(x=3\)

20 tháng 6 2023

\(f\left(x\right)=-3x^2+x-1+x^4-x^3-x^2+3x^4+2x^3\)

\(f\left(x\right)=\left(x^4+3x^4\right)-\left(x^3-2x^3\right)-\left(3x^2+x^2\right)+x-1\)

\(f\left(x\right)=4x^4+x^3-4x^2+x-1\)

\(g\left(x\right)=x^4+x^2-x^3+x-5+5x^3-x^2-3x^4\)

\(g\left(x\right)=\left(x^4-3x^4\right)+\left(5x^3-x^3\right)+\left(x^2-x^2\right)+x-5\)

\(g\left(x\right)=-2x^4+4x^3+x-5\)

`@` `\text {Ans}`

`\downarrow`

`a,`

\(f(x) -3x^2 + x - 1 + x^4 - x^3 - x^2 + 3x^4 + 2x^3\)

`= (x^4 +3x^4) + (-x^3 +2x^3) + (-3x^2 - x^2) + x - 1`

`= 4x^4 + x^3 -4x^2 + x -1`

\(g(x) = x^4 + x^2 - x^3 + x - 5 + 5x^3 - x^2 - 3x^4\)

`= (x^4-3x^4) + (-x^3+5x^3) + (x^2 - x^2) + x -5`

`= -2x^4 + 4x^3 +x - 5`

12 tháng 10 2021

Bài 1

a) \(x=x^5\)

\(x^5-x=0\)

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

\(x=0\) hoặc \(x^4-1=0\)

\(x^4-1=0\)

\(x^4=1\)

\(x=1\)

Vậy x = 0; x = 1

b) \(x^4=x^2\)

\(x^4-x^2=0\)

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

\(x^2=0\) hoặc \(x^2-1=0\)

*) \(x^2=0\)

\(x=0\)

*) \(x^2-1=0\)

\(x^2=1\)

\(x=1\)

Vậy \(x=0\)\(x=1\)

c) \(\left(x-1\right)^3=x-1\)

\(\left(x-1\right)^3-\left(x-1\right)=0\)

\(\left(x-1\right)\left[\left(x-1\right)^2-1\right]=0\)

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

*) \(x-1=0\)

\(x=1\)

*) \(\left(x-1\right)^2-1=0\)

\(\left(x-1\right)^2=1\)

\(x-1=1\) hoặc \(x-1=-1\)

**) \(x-1=1\)

\(x=2\)

**) \(x-1=-1\)

\(x=0\)

Vậy \(x=0\)\(x=1\)\(x=2\)

 

16 tháng 10 2019

a.  64.4x=45

43.4x=45

4x=45:43

4x=42

x=2

14 tháng 8 2025

14 tháng 10 2018

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