Question

№1. сократите дробь:
а)
б)
№2. освободитесь от знака корня в знаменателе:
а)
б)

Answer 1

(5+√a)/(5-√a)*(5+√a)= 1/(5-√a)

√7(√7+1)/√2(√7+1)= √7/√2

18/√6= 18√6/√6*√6= 18√6/6= 3√6

3/(√11+√2)= 3(√11-√2)/(√11+√2)*(√11-√2)= 3(√11-√2)/(√11)²-(√2)²= 3(√11-√2)/(11-2)= 3(√11-√2)/9= (√11-√2)/3

Answer 2

$\frac{5+\sqrt{a}}{25-a}=\frac{5+\sqrt{a}}{5^2-(\sqrt{a})^2}=\frac{5+\sqrt{a}}{(5+\sqrt{a})(5-\sqrt{a})}=\frac{1}{5-\sqrt{a}}\\\\\frac{7+\sqrt{7}}{\sqrt{14}+\sqrt{2}}=\frac{(\sqrt{7})^2+\sqrt{7}}{\sqrt{2}\sqrt{7}+\sqrt{2}}=\frac{\sqrt{7}(\sqrt{7}+1)}{\sqrt{2}(\sqrt{7}+1)}=\frac{\sqrt{7}}{\sqrt{2}}=\sqrt{\frac{7}{2}}=\sqrt{3,5}$

$\frac{18}{\sqrt{6}}=\frac{18\sqrt{6}}{(\sqrt{6})^2}=\frac{18\sqrt{6}}{6}=3\sqrt{6}\\\\\frac{3}{\sqrt{11}+\sqrt{2}}=\frac{3(\sqrt{11}-\sqrt{2})}{(\sqrt{11}+\sqrt{2})(\sqrt{11}-\sqrt{2})}=\frac{3(\sqrt{11}-\sqrt{2})}{(\sqrt{11})^2-(\sqrt{2})^2}=\frac{3(\sqrt{11}-\sqrt{2})}{11-2}=\\\\=\frac{3(\sqrt{11}-\sqrt{2})}{9}=\frac{\sqrt{11}-\sqrt{2}}{3}$