Question

нужно срочно за ранее спасибо большое​

Answer 1

Ответ:

$2) \ 3(\sqrt{7}+y) \ ;$

$4) \ \dfrac{\sqrt{a}-\sqrt{c}}{2\sqrt{b}} \ ;$

Объяснение:

$2) \ \dfrac{21}{\sqrt{7}-y}+\dfrac{3y^{2}}{y-\sqrt{7}}=\dfrac{21}{\sqrt{7}-y}+\dfrac{3y^{2}}{-\sqrt{7}+y}=\dfrac{21}{\sqrt{7}-y}+\dfrac{3y^{2}}{-(\sqrt{7}-y)}=$

$=\dfrac{21}{\sqrt{7}-y}-\dfrac{3y^{2}}{\sqrt{7}-y}=\dfrac{21-3y^{2}}{\sqrt{7}-y}=\dfrac{3(7-y^{2})}{\sqrt{7}-y}=\dfrac{3((\sqrt{7})^{2}-y^{2})}{\sqrt{7}-y}=$

$=\dfrac{3(\sqrt{7}-y)(\sqrt{7}+y)}{\sqrt{7}-y}=3(\sqrt{7}+y);$

$4) \ \dfrac{15\sqrt{d}}{\sqrt{ab}+\sqrt{bc}}:\dfrac{30\sqrt{d}}{a-c}=\dfrac{15\sqrt{d}}{\sqrt{ab}+\sqrt{bc}} \cdot \dfrac{a-c}{30\sqrt{d}}=\dfrac{15\sqrt{d}(a-c)}{30\sqrt{d}(\sqrt{ab}+\sqrt{bc})}=\dfrac{a-c}{2(\sqrt{ab}+\sqrt{bc})}=$

$=\dfrac{(a-c)(\sqrt{ab}-\sqrt{bc})}{2(\sqrt{ab}+\sqrt{bc})(\sqrt{ab}-\sqrt{bc})}=\dfrac{(a-c)(\sqrt{ab}-\sqrt{bc})}{2((\sqrt{ab})^{2}-(\sqrt{bc})^{2})}=\dfrac{(a-c)(\sqrt{ab}-\sqrt{bc})}{2(ab-bc)}=$

$=\dfrac{(a-c)(\sqrt{ab}-\sqrt{bc})}{2b(a-c)}=\dfrac{\sqrt{ab}-\sqrt{bc}}{2b}=\dfrac{\sqrt{b}(\sqrt{a}-\sqrt{c})}{2(\sqrt{b})^{2}}=\dfrac{\sqrt{a}-\sqrt{c}}{2\sqrt{b}};$