1)
0\\x^2+5x-7x-35>0\\x(x+5)-7(x+5)>0\\(x+5)(x-7)>0\\\left \{ {{x+5>0} \atop {x-7>0}} \right. \left \{ {{x+57}} \right. \left \{ {{x" alt="x^2 - 2x - 35 > 0\\x^2+5x-7x-35>0\\x(x+5)-7(x+5)>0\\(x+5)(x-7)>0\\\left \{ {{x+5>0} \atop {x-7>0}} \right. \left \{ {{x+57}} \right. \left \{ {{x" align="absmiddle" class="latex-formula">
Ответ:
x ∈ (7;+∞) ∪ x ∈ (-∞;-5)
2)
- 16\\x^2" alt="-x^2 > - 16\\x^2" align="absmiddle" class="latex-formula">
x ∈ [0;4)
-4, ~~x" alt="x>-4, ~~x" align="absmiddle" class="latex-formula">
Ответ:
x ∈ (-4;4)
3)
-1\\x^2-2x+1>0\\(x-1)^2>0\\(x-1)^2=0\\x=1" alt="x^2-2x>-1\\x^2-2x+1>0\\(x-1)^2>0\\(x-1)^2=0\\x=1" align="absmiddle" class="latex-formula">
x ∈ R \ {1}
Ответ: выражение истино, кроме случая 