Answer:
a) [tex]-\frac{(2\sqrt{3}+5\sqrt{2})^2}{38}[/tex]
b) [tex]\frac{5\sqrt{7}+2 }{171}[/tex]
Step-by-step explanation:
[tex]a) \frac{2\sqrt{3}+5\sqrt{2} }{2\sqrt{3}-5\sqrt{2} }[/tex]
Remember this property;
[tex]\frac{1}{\sqrt{x}+\sqrt{y} }*\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}-\sqrt{y}}{(\sqrt{x})^2-(\sqrt{y})^2 } =\frac{\sqrt{x}-\sqrt{y} }{x-y}[/tex]
We multiply by the same denominator with opposite sign, that gives us a result of the same denominator with each term squared which will cause the term to come out of the root.
Now let's do it with your own equation.
[tex]\frac{2\sqrt{3}+5\sqrt{2} }{2\sqrt{3}-5\sqrt{2} }*\frac{2\sqrt{3}+5\sqrt{2}}{2\sqrt{3}+5\sqrt{2}} =\frac{(2\sqrt{3}+5\sqrt{2})^2}{(2\sqrt{3})^2-(5\sqrt{2})^2}= \frac{(2\sqrt{3}+5\sqrt{2})^2}{(4*3)-(25*2)}=\frac{(2\sqrt{3}+5\sqrt{2})^2}{12-50}=\frac{(2\sqrt{3}+5\sqrt{2})^2}{-38}[/tex]
Now let's try with b.
[tex]b)\frac{1}{5\sqrt{7}-2 } *\frac{5\sqrt{7}+2}{5\sqrt{7}+2}=\frac{5\sqrt{7}+2 }{(5\sqrt{7})^2-(2)^2 }=\frac{5\sqrt{7}+2 }{25*7-4}=\frac{5\sqrt{7}+2 }{175-4}=\frac{5\sqrt{7}+2 }{171}[/tex]
