zad. z dwumianem Newton`a

nice88 · Post autor: **nice88** » 14 gru 2006, o 18:58

(3√7 + √2)^5

dzieki za pomoc w rozwiazaniu

Lorek · Post autor: **Lorek** » 14 gru 2006, o 21:08

\(\displaystyle{ ={5\choose 0} (3\sqrt{7})^5+{5\choose 1} (3\sqrt{7})^4\cdot \sqrt{2}+{5\choose 2}(3\sqrt{7})^3\cdot \sqrt{2}^2+{5\choose 3}(3\sqrt{7})^2\cdot \sqrt{2}^3+\\+{5\choose 4}3\sqrt{7}\cdot \sqrt{2}^4 +{5\choose 5}\sqrt{2}^5=11907\sqrt{7}+19845\sqrt{2}+3780\sqrt{7}+1260\sqrt{2}+\\+60\sqrt{7}+4\sqrt{2}=21109\sqrt{2}+15747\sqrt{7}}\)