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

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

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

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

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

=> x - 7 = 0 hoặc x - 7 = + 1

=> x = 7 hoặc x = 8 hoặc x = 6 

lop 1 moi la la do

Toán lớp 6 mới đúng

ai thấy đúng **** cho mk

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

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 \(x=\left\{6;7;8\right\}\)

x vô nghiệm

(x-7)^x+1 - (x-7)^x+11           = 0

(x-7)^x+1 - (x-7)^x+1 . (x-7)¹⁰ = 0

(x-7)^x+1 . [ 1 - (x-7)¹⁰ ]       =0

(x-7)^x+1 = 0 hoặc 1 -(x-7)¹⁰ = 0

+) (x-7)^x+1  =0

         x-7    = 0

           x     = 7

+) 1- ( x-7)¹⁰ = 0

        (x -7 )¹⁰= 1

x-7 =1 hoặc x-7 = -1

x=8 hoặc x =6

vậy x thuộc {6;7;8}

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

=>(x-7)^x+1.[1-(x-7)^10]=0

TH1:(x-7)^x+1=0<=>x-7=0=<=>x=7

TH2:1-(x-7)^10=0<=>(x-7)^10=1\(\Leftrightarrow\int^{x-7=1}_{x-7=-1}\Leftrightarrow x\in\left\{6;8\right\}\)

vậy x E {6;7;8}

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

Answer:

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

\(\Rightarrow x^2-7x+1-x^2+7x+11=0\)

\(\Rightarrow0x+12=0\)

\(\Rightarrow0x=-12\) (Vô lý)

\(\Rightarrow x\in\varnothing\)

Vậy ta chọn đáp án D.