1/3-0.5=1/3-5/10=1/3-1/2=2/6-3/6=-1/6
3cos^2x+10cosx+3 = 0
cosx = t
3t^2+10t+3 = 0
D = 100-4*3*3=100-36=64=8^2
x1 = (-10 + 8) / 2*3 = -1/3
x2 = (-10 - 8) / 2 * 3= -3 // такого косинуса не существует
обратная замена
cosx = -1/3
ответ: arccos(-1/3)+2Пk
1) (m - 3m^1/3)/(m^2/3 -3) = m^1/3 *(m^2/3 - 3) / (m^2/3 - 3 ) = m^1/3
2) (m^1/2 - n^1/2) / (m^1/4 + n^1/4) = (m^1/4 + n^1/4)(m^1/4 - n^1/4) / (m^1/4 + n^1/4) = m^1/4 - n^1/4
3) (x^1/3 - 2x^1/6y^1/6 + y^1/3) / (x^1/2y^1/3 - x^1/3 y^1/2 = (x^1/6 - y^1/6)(x^1/6 - y^1/6) / x^1/3y^1/3(x^1/6 - y^1/6) =(x^1/6 - y^1/6) / x^1/3y^1/3
(2x+3)^2=5y
(3x=2)^2=5y
(2x+3)^2=(3x+2)^2
4x+12x+9=9x^2+12x+4
5x^2-5=0
x=+/-1
y=5; 1/5
(m-6)(m+6)=m^2-6^2=m²-36
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