[Euler-Fermat]

13. $\sqrt{x^2-3\sqrt{2}x+9}+\sqrt{x^2-4\sqrt{2}x+16} = 5$

ให้ $a = \sqrt{x^2-3\sqrt{2}x+9}, b = \sqrt{x^2-4\sqrt{2}x+16}$

$a^2-b^2 = \sqrt{2}x-7$

$a+b = 5$ ได้ $a-b = \frac{\sqrt{2}x-7}{5}$

จะได้ $10a = \sqrt{2}{x}+18 $

$10\sqrt{x^2-3\sqrt{2}x+9} = \sqrt{2}x+18$

$100x^2-300\sqrt{2}x+900 = 2x^2+36\sqrt{2}x+324$

$98x^2 - 336\sqrt{2}x+576 = 0 $

$x^2 - \frac{24\sqrt{2}}{7}x+\frac{576}{98} = 0 $

$(x-\frac{12\sqrt{2}}{7})^2 = 0 $

$\therefore x = \frac{12\sqrt{2}}{7}$

ได้ $7x^2 - 5\sqrt{2}x+3 = 7(\frac{288}{49})-5\sqrt{2}(\frac{12\sqrt{2}}{7})+3 = 27$