Question

[kien le trong]

(x-y+z)²+(x+y-3)²+(z+5)²=0 tìm x,y,z là số nguyên

Answer 1

[Kiệt Nguyễn]

                                  Giải

Ta có : \(\hept{\begin{cases}\left(x-y+z\right)^2\ge0\\\left(x+y-3\right)^2\ge0\\\left(z+5\right)^2\ge0\end{cases}}\)

Mà \(\left(x-y+z\right)^2+\left(x+y-3\right)^2+ \left(z+5\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-y+z\right)^2=0\\\left(x+y-3\right)^2=0\\\left(z+5\right)^2=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x-y+z=0\\x+y-3=0\\z+5=0\end{cases}}\)

\(\Rightarrow z=0-5\Leftrightarrow z=-5\)

\(\Rightarrow\hept{\begin{cases}x+y=0+3=3\\x-y-5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=3\\x-y=5\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\left(3+5\right)\div2=4\\y=3-4=-1\end{cases}}\)

Vậy \(\hept{\begin{cases}z=-5\\x=4\\y=-1\end{cases}}\)

Answer 2

[Y-S Love SSBĐ]

( x - y + z )² + ( x + y - 3 )² + ( z + 5 )² = 0

<=> \(\hept{\begin{cases}\left(x-y+z\right)^2=0\\\left(x+y-3\right)^2=0\\\left(z+5\right)^2=0\end{cases}}\)

<=> \(\hept{\begin{cases}x-y-5=0\\x+y-3=0\\z=-5\end{cases}}\)

<=> \(\hept{\begin{cases}2x-8=0\\x+y-3=0\\z=-5\end{cases}}\)

<=> \(\hept{\begin{cases}x=4\\y=-1\\z=-5\end{cases}}\)

Hk tốt