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 ^_^ !!!

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

1)  \(32< 2^n< 128\)

\(\Rightarrow2^5< 2^n< 2^7\)

Vì  \(5< n< 7\)

Nên  \(n=6\)

Vậy \(32< 2^6< 128\)

2) \(2.16\ge2^n>4\)
\(\Rightarrow2^5\ge2^n>2^2\)

Vì  \(5\ge n>4\)

nên  \(n=5\)

Vậy   \(2.16\ge2^5>4\)

3/ Tương tự

P/S: chỉ cần đổi các số ra lũy thừa là sẽ tính được!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Kết bạn với mình nha!

a) Ta có: \(\frac{1}{9}\cdot27^n=3^n\)

\(\Leftrightarrow\frac{1}{3^2}\cdot\left(3^3\right)^n=3^n\)

\(\Leftrightarrow3^{3n}=3^{n+2}\)

\(\Rightarrow3n=n+2\)

\(\Rightarrow n=1\)

b) Ta có: \(3^2.3^4.3^n=3^7\)

\(\Rightarrow3^n=3\)

\(\Rightarrow n=1\)

c) Ta có: \(2^{-1}.2^n+4.2^n=9.2^5\)

\(\Leftrightarrow2^n\cdot\frac{9}{2}=9.2^5\)

\(\Rightarrow2^n=2^6\)

\(\Rightarrow n=6\)

d) Ta có: \(32^{-n}.16^n=2048\)

\(\Leftrightarrow\frac{1}{2^{5n}}\cdot2^{4n}=2^{11}\)

\(\Leftrightarrow2^{4n}=2^{5n+11}\)

\(\Rightarrow4n=5n+11\)

\(\Rightarrow n=-11\)

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

Bài 1:

a, \(\left(x-2\right)^2=9\)

\(\Rightarrow x-2\in\left\{-3;3\right\}\Rightarrow x\in\left\{-1;5\right\}\)

b, \(\left(3x-1\right)^3=-8\)

\(\Rightarrow3x-1=-2\Rightarrow3x=-1\)

\(\Rightarrow x=-\dfrac{1}{3}\)

c, \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)

\(\Rightarrow x+\dfrac{1}{2}\in\left\{-\dfrac{1}{4};\dfrac{1}{4}\right\}\)

\(\Rightarrow x\in\left\{-\dfrac{3}{4};-\dfrac{1}{4}\right\}\)

d, \(\left(\dfrac{2}{3}\right)^x=\dfrac{4}{9}\)

\(\Rightarrow\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^2\)

Vì \(\dfrac{2}{3}\ne\pm1;\dfrac{2}{3}\ne0\) nên \(x=2\)

e, \(\left(\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{16}\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^{x-1}=\left(\dfrac{1}{2}\right)^4\)

Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(x-1=4\Rightarrow x=5\)

\(a=lim\dfrac{\left(\dfrac{2}{6}\right)^n+1-\dfrac{1}{4}\left(\dfrac{4}{6}\right)^n}{\left(\dfrac{3}{6}\right)^n+6}=\dfrac{1}{6}\)

\(b=\lim\dfrac{\left(n+1\right)^2}{3n^2+4}=\lim\dfrac{n^2+2n+1}{3n^2+4}=\lim\dfrac{1+\dfrac{2}{n}+\dfrac{1}{n^2}}{3+\dfrac{4}{n^2}}=\dfrac{1}{3}\)

\(c=\lim\dfrac{n\left(n+1\right)}{2\left(n^2-3\right)}=\lim\dfrac{n^2+n}{2n^2-6}=\lim\dfrac{1+\dfrac{1}{n}}{2-\dfrac{6}{n^2}}=\dfrac{1}{2}\)

\(d=\lim\left[1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right]=\lim\left[1-\dfrac{1}{n+1}\right]=1\)

\(e=\lim\dfrac{1}{2}\left[1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right]\)

\(=\lim\dfrac{1}{2}\left[1-\dfrac{1}{2n+1}\right]=\dfrac{1}{2}\)