2^-1.2ⁿ+4.2ⁿ=9.2⁵

=>2^n-1+2².2ⁿ=9.2⁵

=>2^n-1+2ⁿ⁺²=9.2⁵

=>2^n-1.(2³+1)=9.2⁵

=>2^n-1.9=9.2⁵

=>2^n-1=2⁵

=>n-1=5=>n=6

Ta có: \(2^{-1}\cdot2^n+4\cdot2^n=9\cdot2^5\)

\(\Leftrightarrow2^n\cdot2^{-1}+2^n\cdot2^2=9\cdot2^5\)

\(\Leftrightarrow2^n\cdot\left(2^{-1}+2^2\right)=9\cdot2^5\)

\(\Leftrightarrow2^n\cdot\dfrac{9}{2}=9\cdot2^5\)

\(\Leftrightarrow2^n=9\cdot2^5:\dfrac{9}{2}=2^5\cdot9\cdot\dfrac{2}{9}=2^6\)

hay n=6

Vậy: n=6

a) \(\frac{1}{9}.27^n=3^n\)

\(=>\frac{27^n}{9}=3^n\)

\(=>3^n=3^n=>n=1\)

b) \(2^{-1}.2^n+4.2^n=9.2^5\)

\(=>2^{n-1}.2^2.2^n=9.2^5\)

\(=>2^{n-1}.2^{2+n}=9.2^5\)

\(=>2^{2n+1}=9.5^2\)

\(=>n=\)

Câu b đề sai hay sao ấy số xấu lắm

2^-1.2^x+4.2^x=9.2^5

0,5.2^x+4.2^x=288

2^x.(4+0,5)=288

2^x.4,5=288

2^x=288:4,5

2^x=64

2^x=2^6

=) x=6

\(2^{-1}.2^x+4.2^x=9.2^5\)

\(\frac{1}{2}.2^x+4.2^x=9.2^5\)

\(2^x\left(\frac{1}{2}+4\right)=9.2^5\)

\(2^x.\frac{9}{2}=288\)

\(2^x=288:\frac{9}{2}\)

\(2^x=64\)

\(2^x=2^6\)

\(\Rightarrow x=6\)

1/2*2^x+4*2^x=9*25

(1/2+4)*2^x=225

9/2*2^x=225

2^x=225:9/2

2^x=50

=>x\(\in\phi\)

a, 3600: [ ( 5x + 35 ) : x ] = 50

<=> ( 5x + 35 ) : x = 72

<=> 5x + 35 = 72x

<=> 72x - 5x = 35

<=> 67x = 35

<=> x = 35/67

b, 2^n-1 + 4 . 2ⁿ = 9 . 2⁵ 

<=> 2ⁿ : 2 + 4 . 2ⁿ  = 9 . 2⁵ 

<=> 2ⁿ . 1/2 + 4 . 2ⁿ  = 9 . 2⁵ 

<=> 2ⁿ . ( 1/2 + 4 ) = 9 . 2⁵   

<=> 2ⁿ . 9/2 = 9 . 2⁵ 

<=> 2ⁿ : 2⁵ = 9 : 9/2

<=> 2^{n - 5} = 2¹ 

<=> n - 5 = 1

<=> n = 6

a, \(3^{-2}.3^4.3^x=3^7\)

\(\Rightarrow3^{-2+4+x}=3^7\)

\(\Rightarrow3^{2+x}=3^7\)

Vì \(3\ne\pm1;3\ne0\) nên \(2+x=7\Rightarrow x=5\)

b, \(2^{-1}.2^x+4.2^x=9.2^5\)

\(\Rightarrow2^x\left(2^{-1}+4\right)=288\)

\(\Rightarrow2^x.4,5=288\Rightarrow2^x=64=2^6\)

Vì \(2\ne\pm1;2\ne0\) nên \(x=6\)

Chúc bạn học tốt!!!

a)1/9.27^n=3^n

             3^n=3^n

         =>n={0;1;2;3;...}     

a) n= 2;3;5;7;...(n là số nguyên)