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)$frac{sqrt{24}-sqrt{3}}{2sqrt{3}}=frac{2sqrt{6}-sqrt{3}}{2sqrt{3}}=frac{2 sqrt{3}*(sqrt{2}-1)}{2sqrt{3}}=sqrt{2}-1$
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)$frac{2sqrt{b}+2sqrt{3}}{3-b}=frac{2*(sqrt{b}+sqrt{3})}{(sqrt{3}-sqrt{b})*(sqrt{3}+sqrt{b})}=frac{2}{sqrt{3}-sqrt{b}}$
584.
а)$sqrt{3+2sqrt{2}}=sqrt{2}+1 \ \ sqrt{(1+sqrt{2})^2}=sqrt{2}+1 \ \  1+sqrt{2}=sqrt{2}+1$
Доказано.
б)$sqrt{11+4sqrt{6}}=sqrt{3}+2sqrt{2} \ \ sqrt{(2sqrt{2}+sqrt{3})^2}=sqrt{3}+2sqrt{2} \ \ sqrt{3}+2sqrt{2}=sqrt{3}+2sqrt{2}$
Доказано