Select all operations under which the polynomials

Ответ:

izabelllreyes

01.09.2022

Given : $\frac{2}{\sqrt{13}+\sqrt{11}}$

$\mathrm{Multiply\:by\:the\:conjugate}\:\frac{\sqrt{13}-\sqrt{11}}{\sqrt{13}-\sqrt{11}}$

$=\frac{2\left(\sqrt{13}-\sqrt{11}\right)}{\left(\sqrt{13}+\sqrt{11}\right)\left(\sqrt{13}-\sqrt{11}\right)}$

\left(\sqrt{13}+\sqrt{11}\right)\left(\sqrt{13}-\sqrt{11}\right)

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

$a=\sqrt{13},\:b=\sqrt{11}$

$=\left(\sqrt{13}\right)^2-\left(\sqrt{11}\right)^2$

$\left(\sqrt{13}\right)^2-\left(\sqrt{11}\right)^2=13-11=2$

$=\frac{2\left(\sqrt{13}-\sqrt{11}\right)}{2}$

$\mathrm{Divide\:the\:numbers:}\:\frac{2}{2}=1$

$=\sqrt{13}-\sqrt{11}$

Therefore, correct option is $D.\\\sqrt{13} - \sqrt{11}$

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