Question

1÷2√3+√5 how to find this solution

Answer 1

Answer:

1/2√3+√5= 1×√3/2+√5/1=. √3+√5×2/2 =√3×√10/2=√3+√5.

Answer 2

Answer:

$\dfrac{1}{2\sqrt{3}+\sqrt{5}}=\dfrac{2\sqrt{3}-\sqrt{5}}{7}\quad \left(\mathrm{Decimal:\quad }\:0.17543\dots \right)$

Step-by-step explanation:

$\dfrac{1}{2\sqrt{3}+\sqrt{5}}$

$\gray{\mathrm{Multiply\:by\:the\:conjugate}\:\dfrac{2\sqrt{3}-\sqrt{5}}{2\sqrt{3}-\sqrt{5}}}$

$=\dfrac{1\cdot \left(2\sqrt{3}-\sqrt{5}\right)}{\left(2\sqrt{3}+\sqrt{5}\right)\left(2\sqrt{3}-\sqrt{5}\right)}$

$\gray{1\cdot \left(2\sqrt{3}-\sqrt{5}\right)=2\sqrt{3}-\sqrt{5}}$

$\left(2\sqrt{3}+\sqrt{5}\right)\left(2\sqrt{3}-\sqrt{5}\right)$

$\gray{\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}\left(a+b\right)\left(a-b\right)=a^2-b^2}$

$\gray{a=2\sqrt{3},\:b=\sqrt{5}}$

$=\left(2\sqrt{3}\right)^2-\left(\sqrt{5}\right)^2$

$\black{\mathrm{Simplify}\:\left(2\sqrt{3}\right)^2-\left(\sqrt{5}\right)^2}$

$\left(2\sqrt{3}\right)^2-\left(\sqrt{5}\right)^2$

$\gray{\left(2\sqrt{3}\right)^2=12}$

$\gray{\left(\sqrt{5}\right)^2=5}$

$=12-5$

$\gray{\mathrm{Subtract\:the\:numbers:}\:12-5=7}$

$=7$

$=\dfrac{2\sqrt{3}-\sqrt{5}}{7}$