\(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)

\(2^x+2^x.2+2^x.4+2^x.8=480\)

\(2^x.\left(1+2+4+8\right)=480\)

\(2^x.15=480\)

\(2^x=32\)

\(2^x=2^5\)

\(x=5\)

Vậy \(x=5\)

2^x+2^x+1+2^x+2+2^x+3=480

\(\Leftrightarrow2^x\left(1+2+2^2+2^3\right)=480\)

\(\Leftrightarrow2^x.15=480\)

\(\Leftrightarrow2^x=32=2^5\)

\(\Leftrightarrow x=5\)

2^x+2^x+1+2^x+2+2^x+3=480

<=>2^x.(1+2+2²+2³)=480

<=>2^x.15=480

<=>2^x=32=2⁵

<=>x=5

ta co1;2^x+2^x+1+2^x+2+2^x+3=480

=>2^x.1+2^x.2+2^x.2^2+2^x.2^3=480

=>2^x.(1+2+2^2+2^3)=480

=>2^x=480:15=32

=>2^x=2^5

=>x=5

2^x +2^x+1+2^x+2+2^x+3=480

2^x+2^x.2¹+2^x.2²+2^x.2³ = 480

2^x(1+2¹+2²+2³) = 480

2^x.15 = 480

2^x = 480:15

2^x = 32

2^x = 2⁵

=>x=5

\(2^x+2^{x+1}+2^{x+2}+2^{x+3}-480=0\)

\(\Rightarrow2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)

\(\Rightarrow2^x.1+2^x.2+2^x.2^2+2^x.2^3=480\)

\(\Rightarrow2^x\left(1+2+2^2+2^3\right)=480\)

\(\Rightarrow2^x.15=480\)

\(\Rightarrow2^x=480\div15\)

\(\Rightarrow2^x=32=2^5\)

Vậy x = 5

Ta có : \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)

\(\Leftrightarrow2^x\cdot\left(1+2^1+2^2+2^3\right)=480\)

\(\Leftrightarrow2^x\cdot15=480\)

\(\Leftrightarrow2^x=32=2^5\)

\(\Leftrightarrow x=5\)

Vậy : x=5

Ta có : 2^x + 2^{x + 1} + 2^{x + 2} + 2^{x + 3} = 480

=> 2^x.(1 + 2 + 2² + 2³) = 480

=> 2^x.= 32

=> 2^x = 2⁵

=> x = 5

Vậy x = 5

Cỏ giải nè:

2^x + 2^(x + 1) + 2^(x + 2) + 2^(x + 3) - 480 = 0
=>2^x + 2^(x + 1) + 2^(x + 2) + 2^(x + 3) = 480 + 0  
=> 2^x + 2 * 2^x + 2² * 2^x + 2³ * 2^x = 480 
=> 2^x * (1 + 2 + 4 + 8) = 2^x * 15 = 480 
=> 2^x = 32 
=> x = 5 

Không chắc 

<=>  2^x + 2^x.2 + 2^x. 2²+ 2^x.2³ = 480 

<=> 2^x.( 1 + 2+2²+ 2 ³) = 480 

 rồi tính trong ngoặc , lấy 480 chia cho cái kquả đó , rồi phân h kquả vừa chia ra lũy thừa của 2 

\(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)

=> \(2^x.\left(1+2+2^2+2^3\right)=480\)

=>2^x.(1+2+4+8)=480

=>2^x.15=480

=>2^x=480:15

=>2^x=32

=>2^x=2⁵

Vậy x=5.

thu huyền à: 2 dòng trên tui viết bằng FX nên ko copy đc đâu haha

2^x+2^x.2+2^x.4+2^x.8=480

2^x.(1+2+4+8)=480

2^x.15=480

2^x=32

2^x=2^5

suy ra x=5

a) 2^x+2^x+1+2^x+2+2^x+3=480

<=> \(2^x+2^x.2+2^x.2^2+2^x.2^3=480\)

<=> \(2^x.\left(1+2+2^2+2^3\right)=480\)

<=>\(2^x=\frac{480}{1+2+2^2+2^3}=32\)

=> x=5

b) (x²-49)*(x²-81)<0 Khi \(\hept{\begin{cases}x^2-49< 0\\x^2-81>0\end{cases}}\) hoặc \(\hept{\begin{cases}x^2-49>0\\x^2-81< 0\end{cases}}\)

TH1 \(\hept{\begin{cases}x^2-49< 0\\x^2-81>0\end{cases}}\)\(\Rightarrow81< x^2< 49\)(Vô lí)

TH2\(\hept{\begin{cases}x^2-49>0\\x^2-81< 0\end{cases}}\) \(\Rightarrow49< x^2< 81\)\(\Leftrightarrow7^2< x^2< 9^2\)Mà x nguyên \(\Rightarrow x=8\)

c) Làm giống câu a

thank

\(2^x+2^{x+1}+2^{x+2}+2^{x+3}-480=0\)

=> \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)

=>\(2^x+2^x.2+2^x.2^2+2^x.2^3=480\)

=> \(2^x.\left(1+1.2+1.2^2+1.2^3\right)=480\)

=>\(2^x.15=480\)

=> \(2^x=32\)

=>\(2^x=2^5\)

=>\(x=5\)

Vậy x=5

ĐÂY LÀ LỚP 5 À CHÚ