\(\Leftrightarrow\left(x-7\right)\left(x-6\right)\left(x-8\right)=0\)

hay \(x\in\left\{6;7;8\right\}\)

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

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

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

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

Vậy \(x=\left\{6;7;8\right\}\)

\(\Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)

đề sai

x thuộc rỗng nha em 

x rỗng

x thuộc rỗng

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

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

\(\Rightarrow\left(x-7\right)^{x+1}\left[1^2-\left(x-7\right)^{5^2}\right]=0\)

\(\Rightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^5\right]\left[1+\left(x-7\right)^5\right]=0\)

=>(x-7)^x+1=0 hoặc 1-(x-7)⁵=0 hoặc 1+(x-7)⁵=0

+)Nếu (x-7)^x+1=0

=>x-7=0

=>x=7

+)Nếu 1-(x-7)⁵=0

=>(x-7)⁵=1

=>x-7=1

=>x=8

+)Nếu 1+(x-7)⁵=0

=>(x-7)⁵=-1

Vì \(\left(x-7\right)^5\ge0\) với mọi x

=>không tìm được x thỏa mãn 1+(x-7)⁵=0

Vậy x=7 hoặc x=8

x=7 hoặc 8

\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\frac{-14}{5}\right|\)

\(\left|x-\frac{1}{3}\right|=2\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=2\\x-\frac{1}{3}=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\-\frac{5}{3}\end{cases}}}\)

Vậy...

cau b nx

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

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

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

=> (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

=> (x-7)^x+1(1 - (x-7)¹⁰) = 0

=> (x-7)^x+1 = 0

=> x-7 = 0 => x = 7

hoặc 1 - (x-7)¹⁰ = 0

=> (x-7)¹⁰ = 1

=> x-7 = 1

=> x = 8

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

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

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

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

\(\Rightarrow\orbr{\begin{cases}x=7\\x=8\end{cases}}\)

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

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

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

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

\(\Rightarrow\orbr{\begin{cases}x=7\\x=8\end{cases}}\)

Cái này ms đúng 

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

\(\Rightarrow\orbr{\begin{cases}\left[x-7\right]^{x+1}=0\\1-\left[x-7\right]^{10}=0\end{cases}\Rightarrow}\orbr{\begin{cases}x-7=0\\1-\left[x-7\right]^{10}=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=7\\x-7=1\Rightarrow x=8\end{cases}}\)

( x - 7 ) ^ x + 1 - ( x - 7 ) ^ x + 11 = 0 

=> ( x - 7 ) ^ x + 1 = ( x - 7 ) ^ x + 11 

\(x+1\ne x+11\)

=> x - 7 = 1 hoặc 0

Với x - 7 = 1 thì x = 8

Với x - 7 = 0 thì x = 7 

theo mình nghĩ là

x=8 hoặc 7

chính xác ddaaays nhé