Question

Tìm số nguyên x,y biết(x-5)²³ .(y+4)¹³=0

help me

 

Answer 1

(x-5)²³.(y+4)¹³=0

=>(x-5)²³=0        hoặc             (y+4)¹³=0

    (x-5)²³=0²³                           (y+4)¹³=0¹³

=>x-5=0                                  =>y+4 =0

=>x   =5  \(\in Z\)                  =>y         =-4\(\in Z\)

Vậy \(\left(x,y\right)=\left(5,-4\right)\)

Chúc bn học tốt

Answer 2

Ta có : \(\left(x-5\right)^{23}.\left(y+4\right)^{13}=0\)

=> \(\orbr{\begin{cases}\left(x-5\right)^{23}=0\\\left(y+4\right)^{13}=0\end{cases}}\)

=> \(\orbr{\begin{cases}x-5=0\\y+4=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=5\\y=-4\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=5,y\in Z\\y=-4,x\in Z\end{cases}}\)

Answer 3

\(\left(x-5\right)^{23}.\left(y+4\right)^{13}=0\Rightarrow\orbr{\begin{cases}\left(x-5\right)^{23}=0\\\left(y+4\right)^{13}=0\end{cases}\Rightarrow\orbr{\begin{cases}x-5=0\\y+4=0\end{cases}}}\)\(\orbr{\begin{cases}x=5\\y=-4\end{cases}}\)

Vậy x=5, y=-4

Answer 4

\(\left(x-5\right)^{23}.\left(y+4\right)^{13}=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-5\right)^{23}=0\\\left(y+4\right)^{13}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x+4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=-4\end{cases}}\)

Vậy \(x\in\left\{5;-4\right\}\)

Answer 5

Ta có: \(\left(x-5\right)^{23}.\left(y+4\right)^{13}=0\)

\(\implies\)\(\orbr{\begin{cases}\left(x-5\right)^{23}=0\\\left(y+4\right)^{13}=0\end{cases}}\) 

\(\implies\)\(\orbr{\begin{cases}x-5=0\\y+4=0\end{cases}}\)

\(\implies\)\(\orbr{\begin{cases}x=5\\y=-4\end{cases}}\)

Vậy \(x=5;y=-4\)

Answer 6

\(\left(x-5\right)^{23}\cdot\left(y+4\right)^{13}=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-5\right)^{23}=0\\\left(y+4\right)^{13}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x-5=0\\y+4=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=5\\y=-4\end{cases}}}\)