Question

tìm x\(\in\)z

(x - 7)^x+1 \(-\)(x -7 )^x+11 =0

Answer 1

Ta có : (x - 7)^{x + 1} - (x - 7)^{x + 11} = 0 

=>  (x - 7)^{x + 1} (1 - (x - 7)¹⁰) = 0 

\(\Leftrightarrow\orbr{\begin{cases}\left(x-7\right)^{x+1}=0\\1-\left(x-7\right)^{10}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\\left(x-7\right)^{10}=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=7\\\left(x-7\right)=1;-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=7\\x=6;8\end{cases}}\)

Vậy x = {6;7;8} .