b) ĐK: $x\ge 0$.
$18x^2-18x\sqrt x-17x-8\sqrt x-2=0$
$\Leftrightarrow (18x^2-24x\sqrt x-12x)+(6x\sqrt x-8x-4\sqrt x)+(3x-4\sqrt x-2)=0$
$\Leftrightarrow 6x(3x-4\sqrt x-2)+2\sqrt x(3x-4\sqrt x-2)+(3x-4\sqrt x-2)=0$
$\Leftrightarrow (3x-4\sqrt x-2)(6x+2\sqrt x+1)=0$
$\Leftrightarrow 3x-4\sqrt x-2=0$
$\Leftrightarrow 3(\sqrt x-\dfrac 23)^2=\dfrac{10}3$
$\Leftrightarrow (\sqrt x-\dfrac 23)^2=\dfrac{10}9$
$\Leftrightarrow \sqrt x-\dfrac 23=\dfrac{\pm \sqrt{10}}3$
$\Leftrightarrow \sqrt x=\dfrac{2+\sqrt{10}}3$
$\Leftrightarrow x=\dfrac{14+4\sqrt{10}}9$ (TM)
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