Odpowiedzi

2010-01-20T07:34:13+01:00
Tgt2−√3/3 tgt−1− √3/3 tgt=0

tgt² - 2√3/3tgt-1=0

niech tgt=x

x²-2√3/3x-1=0

Δ=(2√3/3)²+4= 4*3/9+4= 12/9+36/9= 48/9
√Δ= √48/9= 4√3/3

x₁=[ 2√3/3-4√3/3]/2= -2√3/6=-√3/3

x₂= 6√3/6= √3

zatem
tgt= -√3/3 lub tgt=√3