9} \atop {x>\frac{1}{2}}} \right. =>x>9 \\2x^2-19x+9-100=0\\2x^2-19x-91=0\\D=19^2+4*2*91=361+728=1089\\x_{1,2}=\frac{19+-33}{4}\\ \left \{ {{x=13} \atop {x=-\frac{7}{2}} \right. " alt="2)lg((x-9)(2x-1))=lg10^2\\ \left \{ {{x>9} \atop {x>\frac{1}{2}}} \right. =>x>9 \\2x^2-19x+9-100=0\\2x^2-19x-91=0\\D=19^2+4*2*91=361+728=1089\\x_{1,2}=\frac{19+-33}{4}\\ \left \{ {{x=13} \atop {x=-\frac{7}{2}} \right. " align="absmiddle" class="latex-formula">
Ответ: 13, т.к -7/2 не подходит по доп. условию (x>9)
доп. условие
0\\x>2" alt="x-2>0\\x>2" align="absmiddle" class="latex-formula">
18" alt="x>18" align="absmiddle" class="latex-formula">
Ответ: (18;+∞)
4)доп условие
0\\x(x-3)>0 \\\left \{ {{x>3} \atop {x<0}} \right. " alt="x^2-3x>0\\x(x-3)>0 \\\left \{ {{x>3} \atop {x<0}} \right. " align="absmiddle" class="latex-formula">
4" alt="x^2-3x<2^2\\x^2-3x-4<0\\x<-1\\x>4" align="absmiddle" class="latex-formula">
Ответ: (-∞;-1) и (4;+∞)