= 1/4 + 1/8 + 1/2 = 2/8 + 1/8 + 4/8 = 7/8
(x-3)√(x^2+x-2)≥0
(x-3)√((x+2)(x-1))≥0
ОДЗ
x ≠ (-2; 1)
---- -2 ----- 1 ------ 3 ------>
- (ОДЗ) - +
Ответ: [3 ; + ∞)
а) x³+2x²+x+2=x²(x+2)+(x+2)=(x+2)(x²+1)
б)4х-4у+ху-у²= 4(x-y)+y(x-y)=(x-y)(4+y)
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