Решить системы уравнений! С решением и пояснением. Буду очень благодарна!

1)
$\left \{ {{-1- \sqrt{17}\ \textless \ x\ \textless \ \sqrt{17}, x \neq\sqrt{17}-1 } \atop {7\sqrt{17}+(\sqrt{17}-7)x-x^2 \geq \sqrt{17}-x}} \right. \\ \\ \left \{ {{-1- \sqrt{17}\ \textless \ x\ \textless \ \sqrt{17}, x \neq\sqrt{17}-1 } \atop {x^2+(6-\sqrt{17})x-6\sqrt{17} \leq 0}} \right.$
$x^2+(6-\sqrt{17})x-6\sqrt{17} = 0$
По теореме Виета:
$\left[\begin{array}{ccc}x_1=-6\\x_2= \sqrt{17} \end{array}$
$\left \{ {{-1- \sqrt{17}\ \textless \ x\ \textless \ \sqrt{17}, x \neq\sqrt{17}-1 } \atop {-6 \leq x \leq \sqrt{17}}} \right$
x ∈ (-1 - √17; √17 - 1)∪(√17 - 1; √17)
2)
$\left \{ {{-7 \textless \ x\ \textless \ -1-\sqrt{17}} \atop {7\sqrt{17}+(\sqrt{17}-7)x-x^2 \leq \sqrt{17}-x}} \right. \\ \\ \left \{ {{-7\ \textless \ x\ \textless \ -1-\sqrt{17}} \atop {x^2+(6-\sqrt{17})x-6\sqrt{17} \geq 0}} \right$
$x^2+(6-\sqrt{17})x-6\sqrt{17} = 0$
По теореме Виета:
$\left[\begin{array}{ccc}x_1=-6\\x_2= \sqrt{17} \end{array}$
$\left \{ {{-7\ \textless \ x\ \textless \ -1-\sqrt{17}} \atop {x \leq -6,x \geq \sqrt{17} \right$
x ∈ (-7; -6]