Question

Освободитесь от иррациональности в знаменателе, 8кл.:

1) 1 / (√5+√3+1)

2) 1 / ( √7+√6+2)

Answer 1

$\frac{1}{ \sqrt{5}+\sqrt{3}+1} =\frac{\sqrt{3}+1-\sqrt{5}}{(\sqrt{3}+1)^2-5}=\frac{\sqrt{3}+1-\sqrt{5}}{3+2\sqrt{3}+1-5}=\frac{\sqrt{3}+1-\sqrt{5}}{2\sqrt{3}-1}=\frac{(\sqrt{3}+1-\sqrt{5})(2\sqrt{3}+1)}{12-1}=$$\frac{6+\sqrt{3}+2\sqrt{3}+1-2\sqrt{15}-\sqrt5}{11}=\frac{7+3\sqrt{3}-2\sqrt{15}-\sqrt5}{11}$

$\frac{1}{ \sqrt{7}+\sqrt{6}+2} =\frac{\sqrt{6}+2-\sqrt{7}}{(\sqrt{6}+2)^2-7}=\frac{\sqrt{6}+2-\sqrt{7}}{6+4\sqrt{6}+4-7}=\frac{\sqrt{6}+2-\sqrt{7}}{4\sqrt{6}+3}=\frac{(\sqrt{6}+2-\sqrt{7})(4\sqrt{6}-3)}{96-9}=$$\frac{(\sqrt{6}+2-\sqrt{7})(4\sqrt{6}-3)}{96-9}=\frac{24-3\sqrt{6}+8\sqrt{6}-6-4\sqrt{42}+3\sqrt{7}}{87}=\frac{18+5\sqrt{6}-4\sqrt{42}+3\sqrt{7}}{87}$