Exercice 14 --- (id : 774)
Calcul dans IR: Exercice 14
correction
a) (52)2(\sqrt 5-2)^2 =(5)22×5×2+22=(\sqrt 5)^2-2\times\sqrt5\times2+2^2 =545+4=5-4\sqrt 5+4 =945=9-4\sqrt 5
Donc 945\sqrt{9-4\sqrt 5} =(52)2=\sqrt{(\sqrt 5-2)^2} =52=52=|\sqrt 5-2|=\sqrt 5-2
b) (35)2(3-\sqrt5)^2 =322×3×5+(5)2=3^2-2\times3\times\sqrt5+(\sqrt5)^2 =965+5=1465=9-6\sqrt5+5=14-6\sqrt5
0<1465<15650<14-6\sqrt5<15-6\sqrt5 donc 1465<1565\sqrt{14-6\sqrt5}<\sqrt{15-6\sqrt5}
Or 1465=(35)2\sqrt{14-6\sqrt5}=\sqrt{(3-\sqrt5)^2} =35=35=|3-\sqrt5|=3-\sqrt5
D'où    35<15653-\sqrt5<\sqrt{15-6\sqrt5}
c) 525=5(52)5-2\sqrt5=\sqrt5(\sqrt5-2) et 45205=5(945)\sqrt{45-20\sqrt5}=\sqrt{5(9-4\sqrt5)} =5945=\sqrt5\sqrt{9-4\sqrt5}
52=945\sqrt5-2=\sqrt{9-4\sqrt5}  donc   525=5(52)5-2\sqrt5=\sqrt5(\sqrt5-2) =5945=\sqrt5\sqrt{9-4\sqrt5} =45205=\sqrt{45-20\sqrt5}