Zad. 1 Zadanie (wyrażenie logiczne): (p v q v r) => {~ p => [(q v r) ^ ~ p] jest prawem rachunku zadań (tautologią). Patrz załącznik Zad. 2 [latex]frac{sqrt{11+ 4sqrt{7}} + sqrt{11- 4sqrt{7}}}{sqrt{11+ 4sqrt{7}} - sqrt{11- 4sqrt{7}}} =frac{sqrt{(2+ sqrt{7})^2} + sqrt{(2- sqrt{7})^2}}{sqrt{(2+ sqrt{7})^2} - sqrt{(2- sqrt{7})^2}} =[/latex] [latex]=frac{|2+ sqrt{7}| + |2- sqrt{7}|}{|2+ sqrt{7}| - |2- sqrt{7}|} =frac{(2+ sqrt{7}) + (sqrt{7}-2)}{(2+ sqrt{7}) - (sqrt{7}-2)} =frac{2+ sqrt{7}+ sqrt{7}-2}{2+ sqrt{7} - sqrt{7}+2} =frac{2sqrt{7}}{4}=frac{sqrt{7}}{2}[/latex] ------------------------------------------------------------------ (2 + √7)² = 2² + 2·2·√7 + (√7)² = 4 + 4√7 + 7 = 11 + 4√7 (2 - √7)² = 2² - 2·2·√7 + (√7)² = 4 - 4√7 + 7 = 11 - 4√7 Zad. 3 [latex][12frac{5}{8} + (frac{1}{3})^{-2} cdot (2 cdot 3^{-1} - 9^{-frac{1}{2}})]^{-frac{1}{3}} = [12frac{5}{8} + 3^2 cdot (2 cdot frac{1}{3} - (frac{1}{9})^{frac{1}{2}})]^{-frac{1}{3}} =[/latex] [latex]=[12frac{5}{8} + 9 cdot (frac{2}{3} - sqrt{frac{1}{9}})]^{-frac{1}{3}} = [12frac{5}{8} + 9 cdot (frac{2}{3} - frac{1}{3})]^{-frac{1}{3}} =(12frac{5}{8} + 9 cdot frac{1}{3})^{-frac{1}{3}} =[/latex] [latex]=(12frac{5}{8} +3)^{-frac{1}{3}} =(15frac{5}{8})^{-frac{1}{3}} = (frac{125}{8})^{-frac{1}{3}} = (frac{8}{125})^{frac{1}{3}} = sqrt[3]{frac{8}{125}} = frac{2}{5}[/latex]
