Ответ: b1) 15 c1) x ∈ (
;+∞)
Объяснение:
b1) 
d∈[-3.25; 7)
S = -3 + -2 + -1 + 0 + 1 + 2 + 3 + 4 + 5 + 6 = 15
c1)
0}}\right.\\\left\{{{9x^{2}-17<9x^{2}-12x+4+15x}\atop{\frac{8x-6}{6}+\frac{3x-9}{6}>0}}\right.\\\left\{{{-17<4+3x}\atop{8x-6+3x-9>0}}\right.\\\left\{{{3x>-21}\atop{11x-15>0}}\right.\\\left\{{{x>-7}\atop{x>\frac{15}{11}}}\right." alt="\left\{{{9x^{2}-17<(3x-2)^{2}+15x}\atop{\frac{4x-3}{3}+\frac{x-3}{2}>0}}\right.\\\left\{{{9x^{2}-17<9x^{2}-12x+4+15x}\atop{\frac{8x-6}{6}+\frac{3x-9}{6}>0}}\right.\\\left\{{{-17<4+3x}\atop{8x-6+3x-9>0}}\right.\\\left\{{{3x>-21}\atop{11x-15>0}}\right.\\\left\{{{x>-7}\atop{x>\frac{15}{11}}}\right." align="absmiddle" class="latex-formula">
x ∈ (
;+∞)