Odpowiedź:
Szczegółowe wyjaśnienie:
a = sin 120° + cos 150° - 2*tg 45° = sin(90°+30°) + cos(90°+60°) - 2*1 (bo tg 45°=1) = cos 30° - sin 60° - 2 = √3/2 - √3/2 - 2 = -2 -> wymierna
b = [(sin 60°-2*cos 135°)/tg 120°]² = [(√3/2 - 2*cos(90°+45°))/tg(90°+30°)]² =
= [(√3/2 + 2*sin(45°))/(-ctg 30°)]² = [(√3/2 - √2)/(-√3)]² = [(√3-2√2)/2√3]² =
[(3 - 2√6)/6]² = (9 - 12√6 + 24)/36 = (33 - 12√6)/36 =
= (11 - 4√6)/12 -> niewymierna
c = (cos 150° + sin 135°)*(cos 30° + sin 45°) = [cos(90°+60°) + sin(90°+45°)]*
[cos 30° + sin 45°] = (-sin 60° + cos 45°)*(cos 30° + sin 45°) =
= (-√3/2 + √2/2)*(√3/2 + √2/2) = -(√3/2 - √2/2)*(√3/2 + √2/2) =
= - (3/4 - 2/4) = - 1/4 -> wymierna
d = tg 150° * cos 30° + sin 90° = tg(90°+60°)* cos 30° + 1 =
= -ctg 60° * cos 30° + 1 = -(√3/3) * √3/2 + 1 = -3/6 + 1 =
= -1/2 + 1 = 1/2 -> wymierna
