What would be the answer to 6 over the square root of 5 minus the square root of 2?
+118608 What would be the answer to 6 over the square root of 5 minus the square root of 2?
I think you mean the simplification rather than the answer:)
\(\frac{6}{\sqrt5}-\sqrt 2\\ =\frac{6}{\sqrt5}*\frac{\sqrt5}{\sqrt5}-\sqrt 2\\ =\frac{6\sqrt5}{5}-\frac{5\sqrt 2}{5}\\ =\frac{6\sqrt5-5\sqrt 2}{5}\\\)
OR maybe you meant:
\(\frac{6}{\sqrt5-\sqrt2}\\ =\frac{6}{\sqrt5-\sqrt2}\times \frac{\sqrt5+\sqrt2}{\sqrt5+\sqrt2}\\ =\frac{6(\sqrt5+\sqrt2)}{(\sqrt5-\sqrt2)(\sqrt5+\sqrt2)}\\ =\frac{6(\sqrt5+\sqrt2)}{5-2}\\ =2(\sqrt5+\sqrt2)\\\)
