Ta có: \(n^2-9+37⋮n-3\)

\(\Leftrightarrow n-3\in\left\{1;-1;37;-37\right\}\)

hay \(n\in\left\{4;2;40;-34\right\}\)

\(n^2-9+37⋮n-3\\ \Rightarrow\left(n-3\right)\left(n+3\right)+37⋮n-3\)

Vì \(\left(n-3\right)\left(n+3\right)⋮\left(n-3\right)\\ \Rightarrow37⋮\left(n-3\right)\\ \Rightarrow n-3\inƯ\left(37\right)=\left\{\pm1;\pm37\right\}\\ \Rightarrow n\in\left\{-34;2;4;40\right\}\)

khó wa

a/ 34 . 3ⁿ : 9 = 34  => 34 . 3ⁿ = 34 x 9  => 34 . 3ⁿ = 306  => 3ⁿ = 306 : 34  => 3ⁿ = 9  => n = 2

b/ 9 < 3ⁿ < 27  => 3² < 3ⁿ < 3³  => 2 < n < 3  

Mà: n thuộc N  => n không tồn tại

c/ Chữ số tận cùng của 360 là 0

d/ Ta có: A =  1 + 3 + 3² + 3³ + 3⁴ + 3⁵ + 3⁶ 

=> 3A = 3 + 3² + 3³ + 3⁴ + 3⁵ + 3⁶ + 3⁷

=> 3A - A = 2A = (3 + 3² + 3³ + 3⁴ + 3⁵ + 3⁶ + 3⁷) - (1 + 3 + 3² + 3³ + 3⁴ + 3⁵ + 3⁶ ) = 3 + 3² + 3³ + 3⁴ + 3⁵ + 3⁶ + 3⁷ -  1 - 3 - 3² - 3³ - 3⁴ - 3⁵ - 3⁶ 

=> 2A = 3⁷ - 1  => A = (3⁷ - 1) : 2  < 3⁷ - 1 = B

=> A < B

câu 1:

(2⁴ .2 .32 .2³) : ( 2² . 32 . 4 )

=( 2⁴ . 2. 2⁵ . 2³) : ( 2² . 2⁵ . 2² )

= 2¹³ . 2⁹

⁼ 2²²

câu 2 tương tự nha bạn

Ta có: \(3^4.3^n:9=37\Rightarrow3^{4+n}:9=37\)

\(\Rightarrow3^{4+n}=\frac{37}{9}\Rightarrow\)Không tồn tại N t/m

\(\dfrac{\dfrac{2}{7}-\dfrac{2}{9}+\dfrac{2}{37}}{\dfrac{5}{7}-\dfrac{5}{9}+\dfrac{5}{37}}-\dfrac{-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}}{\dfrac{5}{6}-\dfrac{5}{8}+\dfrac{5}{10}}=\dfrac{2\left(\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{37}\right)}{5\left(\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{37}\right)}-\dfrac{-1\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}\right)}{\dfrac{5}{2}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}\right)}\)

\(=\dfrac{2}{5}-\dfrac{-1}{\dfrac{5}{2}}=\dfrac{2}{5}+\dfrac{2}{5}=\dfrac{4}{5}\)