+166 \(\dfrac{\sqrt{3x}-4\sqrt{3}}{\sqrt{x}-\sqrt{2}}=\dfrac{2\sqrt{2x}+\sqrt{2}}{\sqrt{6x}-2\sqrt{3}}\)
+118608 \(\dfrac{\sqrt{3x}-4\sqrt{3}}{\sqrt{x}-\sqrt{2}}=\dfrac{2\sqrt{2x}+\sqrt{2}}{\sqrt{6x}-2\sqrt{3}}\\ firstly \qquad x\ne2\qquad \\ (\sqrt{3x}-4\sqrt{3})(\sqrt{6x}-2\sqrt{3})=(2\sqrt{2x}+\sqrt{2})(\sqrt{x}-\sqrt{2})\\ (\sqrt{18x^2}-2*3\sqrt{x}-4\sqrt{18x}+8*3)=(2x\sqrt{2}-4\sqrt{x}+\sqrt{2x}-2)\\ 3x\sqrt{2}-6\sqrt{x}-12\sqrt{2x}+24=2x\sqrt{2}-4\sqrt{x}+\sqrt{2x}-2\\ x\sqrt{2}-2\sqrt{x}-13\sqrt{2x}+26=0\\ \)
\(x\sqrt{2}-2\sqrt{x}-13\sqrt{2x}+26=0\\ let\;\;x=y^2\\ \sqrt2y^2-(2+13\sqrt2)y+26=0\\ \text{Now you can solve for y (use quadratic formula)}\\ \text{Hence solve for x}\\ \text{Just be suspicious of the answers, with all this squaring and square rooting}\\\text{ there is bound to be some hoax answers.}\)
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