Question

Выполнить 3 примера: 1)чему равно (√18+√72-√12)*(√18+√72+√12) 2) 3*√(1/5)+1/2*√20+√(4/5) 3)вычислите 1/(√7-√6)-3/(√6-√3)-4/(√7+√3)

Answer 1

1) $(\sqrt{18}+\sqrt{72}-\sqrt{12})(\sqrt{18}+\sqrt{72}+\sqrt{12})=\\ =(3\sqrt{2}+6\sqrt{2}-2\sqrt{3})(3\sqrt{2}+6\sqrt{2}+2\sqrt{3})=(9\sqrt{2}-2\sqrt{3})(9\sqrt{2}+2\sqrt{3})\\ =81*2-4*3=162-12=150$

2) $3\sqrt{1/5} +\sqrt{20}/2+\sqrt{4/5}=\frac{3}{\sqrt{5}}+\frac{2\sqrt{5}}{2}+\frac{2}{\sqrt{5}} =\frac{5}{\sqrt{5}}+\sqrt{5} =\sqrt{5} +\sqrt{5} =2\sqrt{5}$

3) $\frac{1}{\sqrt{7}-\sqrt{6}}-\frac{3}{\sqrt{6}-\sqrt{3}}-\frac{4}{\sqrt{7}+\sqrt{3}}=\frac{\sqrt{7}+\sqrt{6}}{7-6}-\frac{3(\sqrt{6}+\sqrt{3})}{6-3}-\frac{4(\sqrt{7}-\sqrt{3})}{7-3}=\\ =\sqrt{7}+\sqrt{6}-(\sqrt{6}+\sqrt{3})-(\sqrt{7}-\sqrt{3})=0$