(ab-1)-(ab+1)-(a-b)=ab-1-ab-1-a+b=-a+b-2
(m-mn)-(n-mn)+(m+n) = m-mn-n+mn+m+n=2m
- x^2 + x ≥ 0
x^2 - x =0
x ( x - 1 ) = 0
x= 0 или x - 1 = 0
x = 1
Рисунок во вложении
Ответ: x∈ [ 0 ; 1 ]
(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!
Решение:
18-0,6х+14=х
х+0,6х=-14-18
1,6х=-32
х=20