Odpowiedzi

2009-11-06T21:43:32+01:00
(2-a√3) (1-√3)=√3
2-√3-a√3 +a*3 = √3
2 - a√3 + 3a = 2√3
3a - a√3 = 2√3-2
a(3-√3) = 2√3-2
a = (2√3-2)/(3-√3) - usuwanie niewymiernosci

(2√3-2)/(3-√3) * 3+√3/3+√3 = (2√3-2)(3+√3)/9-3 = 6√3 + 2*3 - 6 - 2√3/6
6√3+6-6-2√3=4√3/6 = 2√3/3

a = 2√3/3
1 5 1
2009-11-06T21:56:46+01:00

2-2√3 - a√3+a√3*3=√3

2-2√3-a√3+3a=√3

3a-a√3=2√3-2

a(3-√3)=2√3-2

2√3-2
a= ----------
(-√3+3)

2√3-2 * (√3-3)
a= ---------------------
(-√3-3) * ( √3-3)

4√3
a= ----------
6
1 5 1
2009-11-06T22:28:53+01:00
2-2√3-a√3+3a=√3 wszystko potegujemy
4-12-3a²+9a²=3
6a²=11
a²=11/6
a=√11/6