đề bài là gì vậy

3x-7^2012=3x-7^2014

suy ra:3x-3x=-7^2014+7^2012

suy ra 7^2012-7^2014=0(vô lí)

nên ko có x thỏa mãn

(3x-7)^2012=(3x-7)^2014

=> (3x-7=0 hoặc =1

+ 3x-7=0 =>3x=7 =>x= 7/3

+3x-7=1 => 3x=8 => x=8/3
 

vì (3x-7)^2012=(3x-7)^2014

=> (3x-7=0 hoặc =1

+ 3x-7=0 =>3x=7 =>x= 7/3

+3x-7=1 => 3x=8 => x=8/3

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

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

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

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

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

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

\(6x\left(1-3x\right)+9x\left(2x-7\right)+171=0\)

\(\Leftrightarrow6x-18x^2+18x^2-63x+171=0\)

\(\Leftrightarrow-57x=-171\)

\(\Leftrightarrow x=3\)

\(\frac{x+1}{2015}+\frac{x+2}{2014}=\frac{x+3}{2013}+\frac{x+4}{2012}\)

\(\Leftrightarrow\left(\frac{x+1}{2015}+1\right)+\left(\frac{x+2}{2014}+1\right)-\left(\frac{x+3}{2013}+1\right)-\left(\frac{x+4}{2012}+1\right)=0\)

\(\Leftrightarrow\)\(\frac{x+2016}{2015}+\frac{x+2016}{2014}-\frac{x+2016}{2013}+\frac{x+2016}{2012}=0\)

\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)

\(\Leftrightarrow x+2016=0\) ( vì \(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\ne0\) )

\(\Leftrightarrow x=-2016\)

Mình chỉ biết bằng 2015