Giải thích các bước giải:

Ta có:
$x=\sqrt{6-3\sqrt{2+\sqrt3}}-\sqrt{2+\sqrt{2+\sqrt3}}$

$\to x=\sqrt{6-\dfrac3{\sqrt2}\cdot \sqrt{4+2\sqrt3}}-\sqrt{2+\dfrac1{\sqrt2}\cdot \sqrt{4+2\sqrt3}}$

$\to x=\sqrt{6-\dfrac3{\sqrt2}\cdot \sqrt{(\sqrt3+1)^2}}-\sqrt{2+\dfrac1{\sqrt2}\cdot \sqrt{(\sqrt3+1)^2}}$

$\to x=\sqrt{6-\dfrac3{\sqrt2}\cdot (\sqrt3+1)}-\sqrt{2+\dfrac1{\sqrt2}\cdot(\sqrt3+1)}$

$\to x=\sqrt{\dfrac{-3\sqrt{3}-3+6\sqrt{2}}{\sqrt{2}}}-\sqrt{\dfrac{\sqrt{3}+1+2\sqrt{2}}{\sqrt{2}}}$

$\to x^2=\dfrac{-3\sqrt{3}-3+6\sqrt{2}}{\sqrt{2}}+\dfrac{\sqrt{3}+1+2\sqrt{2}}{\sqrt{2}}+2\cdot \sqrt{\dfrac{-3\sqrt{3}-3+6\sqrt{2}}{\sqrt{2}}}\cdot\sqrt{\dfrac{\sqrt{3}+1+2\sqrt{2}}{\sqrt{2}}}$

$\to x^2= -\sqrt{6}-\sqrt{2}+8-2\cdot  \sqrt{2\sqrt3-3}$

$\to x^2= -\sqrt{6}-\sqrt{2}+8-2\cdot \sqrt3\cdot  \sqrt{2-\sqrt3}$

$\to x^2= -\sqrt{6}-\sqrt{2}+8-\sqrt2\cdot \sqrt3\cdot  \sqrt{4-2\sqrt3}$

$\to x^2= -\sqrt{6}-\sqrt{2}+8-\sqrt2\cdot \sqrt3\cdot  \sqrt{(\sqrt3-1)^2}$

$\to x^2= -\sqrt{6}-\sqrt{2}+8-\sqrt2\cdot \sqrt3\cdot  (\sqrt3-1)$

$\to x^2=-4\sqrt{2}+8$

$\to x^2-8=-4\sqrt2$

$\to (x^2-8)^2=(-4\sqrt2)^2$

$\to x^4-16x^2+64=32$

$\to x^4-16x^2+32=0$

Ta có:

$P=\dfrac{x^5-16x^3+2x^2+30x+2}{2x^4-29x^2-3x+67}$

$\to P=\dfrac{x(x^4-16x^2)+2x^2+30x+2}{2(x^4-16x^2)+3x^2-3x+67}$

$\to P=\dfrac{x\cdot (-32)+2x^2+30x+2}{2\cdot (-32)+3x^2-3x+67}$

$\to P=\dfrac{2(x^2-x+1)}{3(x^2-x+1)}$

$\to P=\dfrac23$