Question

tìm \(x\in Z\)biết :

\(a)\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)                               \(b)\left(x+20\right)^{100}+|y+4|=0\)

Answer 1

=> 

=> 

=>  hoặc 

+)  => x = 1

+)  =>  hoặc 

=> x = 2 hoặc x = 0

Vậy x = 1 hoặc x = 2 hoặc x = 0

Answer 2

(x-1)^(x+2)=(x-1)^(x+6)

(x-1)^(x+2)-(x-1)^(x+6)=0

(x-1)^(x+2) . [1-(x-1)^4]=0

=> (x-1)^(x+2)=0 hoặc 1-(x-1)^4=0

    x-1=0                        (x-1)^4=1

    x=1                           x-1=1 hoặc x-1=-1

                                        x=2 hoặc x=0

vậy x  {0;1;2} 

Answer 3

a, \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)

\(\Rightarrow\left(x-1\right)^{x+2}-\left(x-1\right)^{x+6}=0\)

\(\Rightarrow\left(x-1\right)^{x+2}\left(1-\left(x-1\right)^4\right)=0\)

\(\Rightarrow\hept{\begin{cases}x-1=0\\1-\left(x-1\right)^4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\x=2\end{cases}}\)

b, \(\left(x+20\right)^{100}+|y+4|=0\)

\(\Rightarrow\hept{\begin{cases}x+20=0\\y+4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-20\\y=-4\end{cases}}\)