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

\(8^x+6=480\)

\(8^x=474\)

       \(\Rightarrow x\in\varnothing\)

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

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

=> 2^x.15 = 480

=> 2^x = 480 : 15

=> 2^x = 32

=> 2^x = 2⁵

=> x = 5

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

\(2^x\cdot1+2^x\cdot2+2^x\cdot4+2^x\cdot8=480\)

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

\(2^x\cdot15=480\)

\(2^x=32\)

do đó x = 5

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

\(\Rightarrow2^x\times\left(1+2+4+8\right)=480\)

\(\Rightarrow2^x\times15=480\)

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

\(\Rightarrow2^x=32\)

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

câu 1 sai đề :>>>

1) \(\left(x-2\right)^6=\left(x-2\right)^8\)

\(\Leftrightarrow\left(x-2\right)^8-\left(x-2\right)^6=0\)

\(\Leftrightarrow\left(x-2\right)^6.\left[\left(x-2\right)^2-1\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\\left(x-2\right)^2=1\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=2\\x=3,x=1\end{cases}}\)

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

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

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

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

\(\Leftrightarrow x=5\)

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

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

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

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

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

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

Mình đamg cần gấp !

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=0

2^x+2^x​.2+2^x.2²+2^x​.2³=0+480

2^x.(1+2+2²+2³)=480

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

 2^x.15=480

2^x=480:15

2^x=32=2⁵

Vậy x =5

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

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

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

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

\(2^x.15=480\)

\(2^x=480:15\)

\(2^x=32\)

\(2^x=2^5\)

\(\Leftrightarrow x=5\)

Vậy x=5

Chúc bạn học tốt

a, 2\(^3\) . x + 2005\(^0\) . x = 994-15:3+1\(^{2025}\) 

   8 .x + 1 . x = 990

x . [ 8 +1 ] = 990

x . 9 = 990

x = 990 : 9

x = 110

các bạn giúp mình với mình đang vội.

a: \(\left(2^3\right)^{1^{2005}}\cdot x+2005^0\cdot x=9915:3+1^{2025}\)

=>\(8\cdot x+1\cdot x=3305+1\)

=>\(9x=3306\)

=>\(x=\dfrac{3306}{9}=\dfrac{1102}{3}\)

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

=>\(2^x+2^x\cdot2+2^x\cdot4+2^x\cdot8=480\)

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

=>\(2^x\cdot15=480\)

=>\(2^x=32\)

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

=>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