√(-5x^2+9x+18)/((x+1)∗(7-x) )
(-5x^2+9x+18)/((x+1)∗(7-x) ) ≥ 0
Одз: (x+1)∗(7-x) ≠ 0,
х ≠ -1, х ≠ 7
решим -5x^2+9x+18 = 0
D=441, x1 = -1,2 ; x2 = 3
отмечаем полученные корни на координатный луч. разобьём на промежутки и найдем знаки на каждом промежутке.
(-∞; -1,2] знак "+", [-1,2; -1) знак "-",
(-1; 3] знак "+", [3; 7) знак "-", (7; -∞) знак "+",
нам нужен промежуток где знак "+",
значит x ϵ промежуткам (-∞; -1,2],(-1; 3], (7; -∞).
ответ: x ϵ промежуткам (-∞; -1,2],(-1; 3], (7; -∞).
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Must Waste More Time...
√(-5x^2+9x+18)/((x+1)∗(7-x) )
Решаем систему:
{-5x^2+9x+18>=0
{x+1 neg 0
{7-x neg 0
-5x^2+9x+18>=0
D=81-4*(-5)*18=441
x1=-1,2 x2=3
-5(x+1,2)(x-3)>=0 |:(-1)
5(x+1,2)(x-3)<=0
+ - +
[-1,2][3]
[-1,2; 3]
Но, здесь надо учесть , что х+1 не равно 0 и 7-х не равно 0, т.е. х не равно -1 и не равно 7.
Получаем, [-1,2;-1)объединённое(-1;3]