5) $\frac{2x^{2}-9x+17}{\sqrt{2x^{2}-6x+16}+\sqrt{3x-1}}=3-x$
ĐK: $\begin{cases} & 2x^2-6x+16 \geq 0 \\ & 3x-1 \geq 0 \end{cases}$
$\dfrac{(2x^2-6x+16)-(3x-1)}{\sqrt{2x^2-6x+16}+\sqrt{3x-1}}=3-x$
$\iff \sqrt{2x^2-6x+16}-\sqrt{3x-1}=3-x$
$\iff \sqrt{2x^2-6x+16}=3-x+\sqrt{3x-1}$
$\longrightarrow 2x^2-6x+16=9-6x+x^2+3x-1+2(3-x)\sqrt{3x-1}$
$\longrightarrow x^2-3x+8=2(3-x)\sqrt{3x-1}$
$\longrightarrow (x^2-3x+8)^2=4(3-x)^2(3x-1)$
$\longrightarrow (x^2-9x+10)^2=0$
$\longrightarrow x=\dfrac{9+\sqrt{41}}{2}$ v $x=\dfrac{9-\sqrt{41}}{2}$
Thử lại ở điều kiện xác định thì chỉ có $x=\dfrac{9-\sqrt{41}}{2}$ thỏa mãn