a)27ⁿ:3ⁿ=9

   (27:3)ⁿ=9

   9ⁿ=9¹

   n=1

Vậy n=1

b)\(\left(\frac{25}{5}\right)^n=5\)

    \(5^n=5^1\)

    n=1

Vạy n=1

c)\(\left(-\frac{81}{3}\right)^n=-243\)

   \(\left(-27\right)^n=\left(-3\right)^5\)

   \(\left[\left(-3\right)^3\right]^n=\left(-3\right)^5\)

   \(\left(-3\right)^{3n}=\left(-3\right)^5\)

    \(3n=5\)

   \(n=\frac{5}{3}\)

Vậy \(n=\frac{5}{3}\)

d)\(\frac{1}{2}.2^n+4.2^n=9.5^n\)

   \(2^n.\left(\frac{1}{2}+4\right)=9.5^n\)

   \(2^n.\frac{9}{2}=3^2.5^n\)

 Mình làm đc mỗi 1 câu, Thông cảm

7^6+7^5+7^4 chia hết cho 11

= 7^4.2^2+7^4.7+7^4

= 7^4.(2^2+7+1)

= 7^4. 11

Vì tích này có số 11 nên => chia hết cho 7

Bài 3:

a: \(35-12n⋮n\)

\(\Leftrightarrow n\in\left\{1;5;7;35\right\}\)

b: \(n+13⋮n+5\)

\(\Leftrightarrow n+5\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)

hay \(n\in\left\{-4;-6;-3;-7;-1;-9;3;-13\right\}\)

a,3^-1.3ⁿ+6.3^n-1=7.3⁶

=>3^n-1+6.3^n-1=7.3⁶

=>3^n-1.(1+6)=7.3⁶

=>7.3^n-1=7.3⁶

=>n-1=6

=>n=7

a) \(\dfrac{1}{8}.16^n=2^n\)

=>\(\dfrac{2^{4n}}{2^3}=2^n\)

=>\(2^{4n-3}=2^n\)

=>4n-3=n

=>3n-3=0

=>n=1.

b) \(27< 3^n< 243\)

=>\(3^3< 3^n< 3^5\). Mà n là số tự nhiên.

- Vậy n=4

a) \(\dfrac{1}{8}.16^n=2^n\)

\(\Rightarrow2^{4n}=2^3.2^n\)

\(\Rightarrow4n=3+n\)

\(\Rightarrow3n=3\)

\(\Rightarrow n=1\)

Vậy: \(n=1\)

b) \(27< 3^n< 243\)

\(\Rightarrow3^3< 3^n< 3^5\)

\(\Rightarrow3< n< 5\)

\(\Rightarrow n=4\)

Vậy: \(n=4\)

1. Ta có: \(x\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)

=> \(x\left(6-x\right)^{2003}-\left(6-x\right)^{2003}=0\)

=> \(\left(6-x\right)^{2003}\left(x-1\right)=0\)

=> \(\orbr{\begin{cases}\left(6-x\right)^{2003}=0\\x-1=0\end{cases}}\)

=> \(\orbr{\begin{cases}6-x=0\\x=1\end{cases}}\)

=> \(\orbr{\begin{cases}x=6\\x=1\end{cases}}\)

Bài 2. Ta có: (3x - 5)¹⁰⁰ \(\ge\)0 \(\forall\)x

       (2y + 1)¹⁰⁰ \(\ge\)0 \(\forall\)y

=> (3x - 5)¹⁰⁰ + (2y + 1)¹⁰⁰ \(\ge\)0 \(\forall\)x;y

Dấu "=" xảy ra khi: \(\hept{\begin{cases}3x-5=0\\2y+1=0\end{cases}}\) => \(\hept{\begin{cases}3x=5\\2y=-1\end{cases}}\) => \(\hept{\begin{cases}x=\frac{5}{3}\\y=-\frac{1}{2}\end{cases}}\)

Vậy ...

\(a,\frac{16}{2^n}=2\)

\(\Rightarrow\frac{16}{2^n}=\frac{2}{1}\)

\(\Rightarrow2^n.2=16\)

\(\Rightarrow4^n=4^2\)

\(\Rightarrow n=2\)

\(b,\frac{\left(-3\right)^n}{81}=-27\)

\(\Rightarrow\left(-3\right)^n=81.\left(-27\right)\)

\(\Rightarrow\left(-3\right)^n=-2187\)

\(\Rightarrow\left(-3\right)^n=\left(-3\right)^7\)

\(\Rightarrow n=7\)

16/2^n=2 => 2^n .2 =16 => 2^n=8=2^3=>n=3