Question

вправа 582 584 срочно помогите

Answer 1

582.
а)$\frac{\sqrt{14}-\sqrt{2}}{\sqrt{7}-1}=\frac{\sqrt{2}*(\sqrt{7}-1)}{\sqrt{7}-1}=\sqrt{2}$
б)$\frac{\sqrt{24}-\sqrt{3}}{2\sqrt{3}}=\frac{2\sqrt{6}-\sqrt{3}}{2\sqrt{3}}=\frac{2 \sqrt{3}*(\sqrt{2}-1)}{2\sqrt{3}}=\sqrt{2}-1$
в)$\frac{3+\sqrt{3}}{\sqrt{21}+\sqrt{7}}=\frac{\sqrt{3}*(\sqrt{3}+1)}{\sqrt{7}*(\sqrt{3}+1)}= \frac{\sqrt{3}}{\sqrt{7}}$
г)$\frac{x^2-2}{x-\sqrt{2}}=\frac{(x- \sqrt{2})*(x+\sqrt{2})}{x-\sqrt{2}}=x+\sqrt{2}$
д)$\frac{a+\sqrt{5}}{a^2-5}=\frac{a+\sqrt{5}}{(a-\sqrt{5})*(a+\sqrt{5})}=\frac{1}{a-\sqrt{5}}$
е)$\frac{2\sqrt{b}+2\sqrt{3}}{3-b}=\frac{2*(\sqrt{b}+\sqrt{3})}{(\sqrt{3}-\sqrt{b})*(\sqrt{3}+\sqrt{b})}=\frac{2}{\sqrt{3}-\sqrt{b}}$
584.
а)$\sqrt{3+2\sqrt{2}}=\sqrt{2}+1 \\ \\ \sqrt{(1+\sqrt{2})^2}=\sqrt{2}+1 \\ \\  1+\sqrt{2}=\sqrt{2}+1$
Доказано.
б)$\sqrt{11+4\sqrt{6}}=\sqrt{3}+2\sqrt{2} \\ \\ \sqrt{(2\sqrt{2}+\sqrt{3})^2}=\sqrt{3}+2\sqrt{2} \\ \\ \sqrt{3}+2\sqrt{2}=\sqrt{3}+2\sqrt{2}$
Доказано