Question

ПОМОГИТЕ 5 ЗАДАНИЕ, ПЕРЕПУТАЛА

Answer 1

Ответ ниже:

Формулы. Разность квадратов: $a^2-b^2=(a-b)(a+b)$

$\frac{a-4}{\sqrt{a}-2}}=\frac{(\sqrt{a})^2-(2)^2}{\sqrt{a}-2}=\frac{(\sqrt{a}-2)(\sqrt{a}+2)}{\sqrt{a}-2}=\sqrt{a}+2$

$\frac{\sqrt{23}-23}{\sqrt{23}}=\frac{\sqrt{23}}{\sqrt{23}}-\frac{23}{\sqrt{23}}=1-\frac{23*\sqrt{23}}{\sqrt{23}*\sqrt{23}}=1-\frac{23\sqrt{23}}{23}=1-\sqrt{23}$

$\frac{a-5}{a+2\sqrt{5a}+5}=\frac{(\sqrt{a})^2-(\sqrt{5})^2}{(\sqrt{a})^2+2*\sqrt{a}*\sqrt{5}+(\sqrt{5})^2}=\frac{(\sqrt{a}-\sqrt{5})(\sqrt{a}+\sqrt{5})}{(\sqrt{a}+\sqrt{5})^2}=\frac{\sqrt{a}-\sqrt{5}}{\sqrt{a}+\sqrt{5}}=\\=\frac{(\sqrt{a}-\sqrt{5})*(\sqrt{a}-\sqrt{5})}{(\sqrt{a}+\sqrt{5})(\sqrt{a}-\sqrt{5})}=\frac{(\sqrt{a}-\sqrt{5})^2}{a-5}=\frac{a-2\sqrt{5a}+5}{a-5}$