<span>(32)^x=16</span>
<span>(2^5)^x = 2^4</span>
<span>5x = 4</span>
<span>x = 0,8</span>
<span>
</span><span><span>(у^2 -4) * 3/(y-2)^2 = 3(y-2)(y+2)/(y-2)^2 = 3(y+2)/(y-2)</span></span>
2x^2-5x+4=0
D=(-5)^2-4*2*4=25-32=-7 D<0 - нет корней
Y' = –8 + 4/cos2x
(–8·cos2x + 4)/cos2x = 0
–8·cos2x + 4 = 0
cos2x = 1/2
cosx = +– 1/√2
x = 3π/4
x = π/4
о отрезок [– pi/3 ; pi/3] входит только π/4
y(π/4) = – 8·(π/4) + 4·1 + 2pi + 1 = –4π + 2π + 5 = –2π + 5
еще проверить y(– pi/3 ) и y( pi/3)
6*а+3*n=m - математическая модель данной ситуации