Question

Tìm x,y,z

x²-6x+y²+10y+34= -(4z-1)²

Answer 1

\(x^2-6x+y^2+10y+34=-\left(4z-1\right)^2\)

\(\Leftrightarrow\left(x^2-6x+9\right)+\left(y^2+10y+34\right)+\left(4z-1\right)^2=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(y+5\right)^2+\left(4z-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-3\right)^2=0\\\left(y+5\right)^2=0\\\left(4z-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-5\\z=\dfrac{1}{4}\end{matrix}\right.\)

Vậy........