Điều kiện xác định : $-\sqrt[]{17}\le x\le\sqrt[]{17}$
Đặt $t=x+\sqrt[]{17-x^2}(t\ge0)$
$t^2=x^2+2x\sqrt[]{17-x^2}+17-x^2$
$t^2=2x\sqrt[]{17-x^2}+17$
$x\sqrt[]{17-x^2}=\dfrac{t^2-17}{2}$
$PTTT⇔t+\dfrac{t^2-17}{2}=9$
$⇔2t+t^2-17=18$
$⇔t^2+2t-35=0$
$⇔$\(\left[ \begin{array}{l}t=5(n)\\t=-7(l)\end{array} \right.\)
$⇒x+\sqrt[]{17-x^2}=5$
$⇔\sqrt[]{17-x^2}=5-x$
Điều kiện xác định
$5-x≥0⇔x≤5$
$⇒17-x^2=(5-x)^2$
$⇔17-x^2=25-10x+x^2$
$⇔2x^2-10x+8=0$
$⇔$\(\left[ \begin{array}{l}x=1(n)\\x=4(n)\end{array} \right.\)
Vậy $S=\{1;4\}$
