Simplify 3+√5/3-√5-3-√5/3+√5plzz do not send any rubbish answe

1.	Simplify 3+√5/3-√5-3-√5/3+√5plzz do not send any rubbish answersif you don't know then leave it
Answer» Correct question : Simplify : $\frac{3+\sqrt{5}}{3-\sqrt{5}} -\frac{3-\sqrt{5}}{3+\sqrt{5}}$ ANSWER : = 3√5 Step-by-step EXPLANATION : $\underline{\underline{\bf Rationalizing \ factor:}}$ => The factor of multiplication by which rationalization is done, is called as rationalizing factor. => If the product of TWO surds is a RATIONAL number, then each surd is a rationalizing factor to other. => For example, rationalizing factor of (3 + √2) is (3 - √2) ___________________________ $\bf =\frac{3+\sqrt{5}}{3-\sqrt{5}} -\frac{3-\sqrt{5}}{3+\sqrt{5}} \\\\\\ =(\frac{3+\sqrt{5}}{3-\sqrt{5}} \times \frac{3+\sqrt{5}}{3+\sqrt{5}})-(\frac{3-\sqrt{5}}{3+\sqrt{5}}\times \frac{3-\sqrt{5}}{3-\sqrt{5}}) \\\\\\ =\frac{(3+\sqrt{5})^2}{(3-\sqrt{5})(3+\sqrt{5})} -\frac{(3-\sqrt{5})^2}{(3+\sqrt{5})(3-\sqrt{5})} \\\\\\ =\frac{3^2+\sqrt{5}^2+2(3)(\sqrt{5})}{3^2-\sqrt{5}^2} -\frac{3^2+\sqrt{5}^2-2(3)(\sqrt{5})}{3^2-\sqrt{5}^2} \\\\\\ =\frac{9+5+6\sqrt{5}}{9-5} -\frac{9+5-6\sqrt{5}}{9-5} \\\\\\$ $\bf =\frac{14+6\sqrt{5}}{4} -\frac{14-6\sqrt{5}}{4} \\\\\\ =\frac{14+6\sqrt{5}-14+6\sqrt{5}}{4} \\\\\\ =\frac{12\sqrt{5}}{4} \\\\\\ =3\sqrt{5}$ _______________________________ If the question is :- $3+\frac{\sqrt{5}}{3} -\sqrt{5}-3-\frac{\sqrt{5}}{3}+\sqrt{5}$ answer : $= 3-3+\frac{\sqrt{5}}{3}-\frac{\sqrt{5}}{3}-\sqrt{5}+\sqrt{5} \\\\ =0$

Simplify 3+√5/3-√5-3-√5/3+√5plzz do not send any rubbish answersif you don't know then leave it

Answer»

Correct question :

Simplify : $\frac{3+\sqrt{5}}{3-\sqrt{5}} -\frac{3-\sqrt{5}}{3+\sqrt{5}}$

ANSWER :

= 3√5

Step-by-step EXPLANATION :

$\underline{\underline{\bf Rationalizing \ factor:}}$

=> The factor of multiplication by which rationalization is done, is called as rationalizing factor.

=> If the product of TWO surds is a RATIONAL number, then each surd is a rationalizing factor to other.

=> For example, rationalizing factor of (3 + √2) is (3 - √2)

___________________________

$\bf =\frac{3+\sqrt{5}}{3-\sqrt{5}} -\frac{3-\sqrt{5}}{3+\sqrt{5}} \\\\\\ =(\frac{3+\sqrt{5}}{3-\sqrt{5}} \times \frac{3+\sqrt{5}}{3+\sqrt{5}})-(\frac{3-\sqrt{5}}{3+\sqrt{5}}\times \frac{3-\sqrt{5}}{3-\sqrt{5}}) \\\\\\ =\frac{(3+\sqrt{5})^2}{(3-\sqrt{5})(3+\sqrt{5})} -\frac{(3-\sqrt{5})^2}{(3+\sqrt{5})(3-\sqrt{5})} \\\\\\ =\frac{3^2+\sqrt{5}^2+2(3)(\sqrt{5})}{3^2-\sqrt{5}^2} -\frac{3^2+\sqrt{5}^2-2(3)(\sqrt{5})}{3^2-\sqrt{5}^2} \\\\\\ =\frac{9+5+6\sqrt{5}}{9-5} -\frac{9+5-6\sqrt{5}}{9-5} \\\\\\$

$\bf =\frac{14+6\sqrt{5}}{4} -\frac{14-6\sqrt{5}}{4} \\\\\\ =\frac{14+6\sqrt{5}-14+6\sqrt{5}}{4} \\\\\\ =\frac{12\sqrt{5}}{4} \\\\\\ =3\sqrt{5}$

_______________________________

If the question is :-

$3+\frac{\sqrt{5}}{3} -\sqrt{5}-3-\frac{\sqrt{5}}{3}+\sqrt{5}$

answer :

$= 3-3+\frac{\sqrt{5}}{3}-\frac{\sqrt{5}}{3}-\sqrt{5}+\sqrt{5} \\\\ =0$