3x-pi/4=+-pi/3+2pi*n, n€z
3x-pi/4=pi/3+2pi*n
x=(7pi+24pi*n)/36
при n=1:
X=31pi/36
(1/3*1/2m+1/3*(-3)-m*1/2m-m*(-3)
(1/3*1/2m+1*(-1)-m*1/2m-m*(-3)
(m/2*3+1*(-1)-m*1/2m-m*(-3)
1/6m+1*(-a)-m=1/2m-m*(-3)
(1/6m-1-m*1/2m-m*(-3)
(1/6m-1-1/2m^2-m*(-3)
(1/6m-1-1/2m^2+3m)
(-1/2m^2+1/6m+3m-1)
(-1/2m^2+1/6m+3m*6/6-1)
(-1/2m^2+1/6m+18m/6-1)
(-1/2m^2+1/6(m+18m)-1)
(-1/2m^2+19m/6-1)
-1/2m^2+19m/6-1 ответ!
1) (a^2-5a+6)/(a^2+7a+12)*(a^2+3a)/(a^2-4a+4)=((a-3)(a-2))/((a+3)(a+4))*(a(a+3))/((a-2)(a-2))=(a(a-3))/((a-2)(a+4))
2) (x^2+2x-3)/(x^2+3x-10)*(x^2-9x+14)/(x^2+7x+12)=((x-1)(x+3))/((x-2)(x+5))*((x-7)(x-2))/((x+3)(x+4))=(x-1)(x-7)/(x+5)(x+4)
Вроде так, но могу ошибаться)
6^x≤6^1
(6 > 1) x ≤ 1
x-1 < 0
x < 0
3^x≤3^1
(3 > 1) x ≤1
(1/3)^x ≥3
3^(-x) ≥ 3
(3 > 1) -x ≥3
x ≤ - 3
3^x ≥ 3^1
(3 > 1) x ≥ 1