a) 3^2x + 3^2x+1 = 324

3^2x . 1 + 3^2x . 3 = 324

3^2x . ( 1 + 3 ) = 324

3^2x . 4 = 324

3^2x = 324 : 4

3^2x = 81

3^2x = 3⁴

=> 2x = 4

=> x = 4 : 2 = 2

( 3x - 7 )²⁰¹² = ( 3x - 7 )²⁰¹⁴

( 3x - 7 )²⁰¹⁴ - ( 3x - 7 )²⁰¹² = 0

( 3x - 7 )²⁰¹² . [ ( 3x - 7 )² - 1 ] = 0

\(\Rightarrow\orbr{\begin{cases}\left(3x-7\right)^{2012}=0\\\left(3x-7\right)^2-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x-7=0\\\left(3x-7\right)^2=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x=7\\3x-7=1\text{ hoặc }3x-7=-1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\3x=8\text{ hoặc }3x=6\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=\frac{8}{3}\text{ hoặc }x=2\end{cases}}\)

=> (x-7)^x+1  +  1.[1-(x-7)^10] = 0

=> x-7 = 0 hoặc 1-(x-7)^10 = 0

=> x=7 hoặc x = 8 hoặc x = 6 

k mk nha

a.

để x \(\in Z\) thì \(2x-3\inƯ_{\left(7\right)}\left\{1;-1;7;-7\right\}\)

Vậy x ={2;1;5;-2}

câu b mik ko pt lm đâu nhé sorry

c.

\(\frac{3x+2}{x+1}=\frac{3x+3-1}{x+1}=3-\frac{1}{x+1}\)

để x \(\in Z\) thì \(x+1\inƯ_{\left(1\right)}\left\{1;-1\right\}\)

câu d giống câu a rồi nhé

\(\left(\frac{3}{5}\right)^{2x}=\frac{9}{25}\)

\(\left(\frac{3}{5}\right)^{2x}=\frac{3^2}{5^2}\)

\(\left(\frac{3}{5}\right)^{2x}=\left(\frac{3}{5}\right)^2\)

\(\Rightarrow2x=2\)

\(\Rightarrow x=1\)

vậy \(x=1\)

=> 2x-3 thuộc Ư(7)={-1,-7,1,7}

2x-3=-1 => x=1

2x-3=-7 => x=-2

2x-3=1 => x=2

2x-3=7 => x=5

Vậy A=\(\frac{7}{2.1-3}=\frac{7}{-1}=-7\)

\(A=\frac{7}{2.\left(-2\right)-3}=\frac{7}{-7}=-1\)

A=\(\frac{7}{2.2-3}=\frac{7}{1}=7\)

\(A=\frac{7}{2.5-3}=\frac{7}{7}=1\)