\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)

\(\Rightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=7\\\left(x-7\right)^{10}=\left(\pm1\right)^{10}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=7\\x-7=1\\x-7=-1\end{cases}}\)

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

P/S: 2 dòng cuối bn thay \(\hept{\begin{cases}\\\\\end{cases}}\)thành \(\orbr{\begin{cases}\\\end{cases}}\)nha 

(x+5)^x+1=(x+5)^x+11

Ta có: (x+5)^x+11 : {(x+5)^x+1}=1

Suy ra (x+5)^x+10=1

Do (x+5)^x+1 - (x+5)^x+11 = 0 nên:

(x+5)^{x+1 =} (x+5)^x+11.

^{=> Bạn kiểm tra lại đề bài đi ha!}

7^{x + 2} + 2.7^{x - 1} = 345

7^x . 7² + 2 . 7^x : 7 = 345

7^x . 49 + 2 . 7^x . \(\frac{1}{7}\)= 345

7^x . (49 + 2.\(\frac{1}{7}\)) = 345

7^x . \(\frac{345}{7}\)= 345

7^x = 345 . \(\frac{7}{345}\)

7^x = 7

=> x = 1 

tao chịu

\(\left(x-7\right)^{x-1}-\left(x-7\right)^{x+11}=0\)

\(\Rightarrow\left(x-7\right)^{x+1}.\left[1-\left(x-7\right)^{10}\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\1-\left(x-7\right)^{10}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x-7=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0+7\\x-7=1\\x-7=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=7\\x=1+7\\x=\left(-1\right)+7\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=7\\x=8\\x=6\end{matrix}\right.\)

Vậy \(x\in\left\{7;8;6\right\}.\)

Chúc bạn học tốt!