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)

 a)1/9.27^n=3^n

 3^n=3^n

=>n={0;1;2;3...}

 Tích nha ^_^ !!!

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, 9.27ⁿ=3ⁿ

    3².3³ⁿ=3ⁿ

    3²⁺³ⁿ=3ⁿ

     2+3n=n

     n-3n=2

     -2n=2

       n=-1

bạn nhớ k cho mk nha

b, (2³:4).2ⁿ=4

    (2³:2²).2ⁿ=2²

      2¹.2ⁿ=2²

      2¹⁺ⁿ = 2²

       1+n=2

          n=1

bạn nhớ k cho minh nha

vậy c và d đâu rùi bạn

đề bài này rất hay, chị làm dùm a) thui nhe

a) 27ⁿ. 1/9 = 3³ⁿ / 3²

    vậy 3³ⁿ / 3² = 3ⁿ

           3³ⁿ = 3ⁿ⁺²

 n =1

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

\(2^n.\left(\frac{1}{2}.4\right)=288\)

\(2^n.2=288\)

\(2^n=288:2\)

\(2^n=144\)

       Suy ra n ko tìm được

Ta có : 

\(\frac{1}{2}\cdot2^n+4\cdot2^n=\frac{9}{2}\cdot2^5\)

\(=>\left(\frac{1}{2}+4\right)\cdot2^n=9\cdot2^5\)

\(=>\left(\frac{1}{2}+\frac{8}{2}\right)\cdot2^n=9\cdot2^5\)

\(=>\frac{9}{2}\cdot2^n=\frac{9}{2}\cdot2^5\)

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

\(=>n=5\)

\(\frac{1}{2}\times2^n+4\times2^n=9\times2^5\)

\(2^n\times\left(\frac{1}{2}+4\right)=9\times2^5\)

\(2^n\times\left(\frac{1+8}{4}\right)=9\times2^5\)

\(2^n\times\frac{9}{2^2}=9\times2^5\)

\(\frac{2^n}{2^2}=2^5\)

\(2^n=2^5\times2^2\)

\(2^n=2^7\)

\(n=7\)

2ⁿ.(1/2+4)=9.32

2ⁿ*4,5=288

2ⁿ=64=2⁶

=> n=6

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

\(2^n.2=9.2^5\)

\(\Rightarrow2^n=9.2^4\)

Ko có n nhé bn