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$$\dfrac{\sqrt{3x}-4\sqrt{3}}{\sqrt{x}-\sqrt{2}}=\dfrac{2\sqrt{2x}+\sqrt{2}}{\sqrt{6x}-2\sqrt{3}}$$

Creeperhissboom  Apr 3, 2018
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#1
-1

What do you want done with this? Simplify it? Solve for x? .........etc.

Guest Apr 3, 2018
#2
+2404
+2

Is it possible to cross multiply?

tertre  Apr 3, 2018
#3
+92254
+3

$$\dfrac{\sqrt{3x}-4\sqrt{3}}{\sqrt{x}-\sqrt{2}}=\dfrac{2\sqrt{2x}+\sqrt{2}}{\sqrt{6x}-2\sqrt{3}}\\ firstly \qquad x\ne2\qquad \\ (\sqrt{3x}-4\sqrt{3})(\sqrt{6x}-2\sqrt{3})=(2\sqrt{2x}+\sqrt{2})(\sqrt{x}-\sqrt{2})\\ (\sqrt{18x^2}-2*3\sqrt{x}-4\sqrt{18x}+8*3)=(2x\sqrt{2}-4\sqrt{x}+\sqrt{2x}-2)\\ 3x\sqrt{2}-6\sqrt{x}-12\sqrt{2x}+24=2x\sqrt{2}-4\sqrt{x}+\sqrt{2x}-2\\ x\sqrt{2}-2\sqrt{x}-13\sqrt{2x}+26=0\\$$

$$x\sqrt{2}-2\sqrt{x}-13\sqrt{2x}+26=0\\ let\;\;x=y^2\\ \sqrt2y^2-(2+13\sqrt2)y+26=0\\ \text{Now you can solve for y (use quadratic formula)}\\ \text{Hence solve for x}\\ \text{Just be suspicious of the answers, with all this squaring and square rooting}\\\text{ there is bound to be some hoax answers.}$$

Melody  Apr 3, 2018

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