Bài giải:

Ta có:

x² – 6x + y² + 10y + 34

 = x²- 6x + y² + 10y + 3² + 5²

= x² – 2.3.x + 3² + y² + 2.5.y + 5²

= (x – 3²) + (y + 5)²

$\Rightarrow $(x – 3²) + (y + 5)² = -(4z – 1)²

$\Rightarrow $(x – 3²) + (y + 5)² + (4z – 1)² = 0

Mà: (x – 3²) $\ge $ 0  $\forall $x

        (y + 5)² $\ge $ 0  $\forall $y

        (4z – 1)^{2 $\ge $} 0  $\forall $z

$\Rightarrow $ $\left\{ \begin{align}& {{(x-3)}^{2}}=0 \\  & {{(y+5)}^{2}}=0 \\ & {{(4z-1)}^{2}}=0 \\\end{align} \right.$

$\Rightarrow $$\left\{ \begin{align}& x-{{3}^{{}}}=0 \\ & y+{{5}^{{}}}=0 \\ & 4z-{{1}^{{}}}=0 \\ \end{align} \right.$

$\Rightarrow $$\left\{ \begin{align}& x={{3}^{{}}} \\ & y=-{{5}^{{}}} \\ &z={{\frac{1}{4}}^{{}}} \\ \end{align} \right.$

Vậy: y = 5