Решите ** множестве R систему неравенства:(9x+9)(x-4)>9x²+x+5=2x-3-(x-3)≤5=

Question

Решите на множестве R систему неравенства:
(9x+9)(x-4)>9x²+x+5=
2x-3-(x-3)≤5=

Answer 1

9x^2+x+5} \atop {2x-3-(x-3)\leq5}} \right.\\ \left \{ {{9x^2-27x-36 > 9x^2+x+5} \atop {2x-3-x+3\leq5}} \right.\\ \left \{ {{9x^2-27x-36-9x^2-x-5>0} \atop {x\leq5}} \right.\\ \left \{ {{-28x-41>0} \atop {x\leq5}} \right.\\ \left \{ {{-28x>41} \atop {x\leq5}} \right.\\ \left \{ {{x<-\frac{41}{28}} \atop {x\leq5}} \right.\\ x<-\frac{41}{28} " alt=" \left \{ {{(9x+9)(x-4) > 9x^2+x+5} \atop {2x-3-(x-3)\leq5}} \right.\\ \left \{ {{9x^2-27x-36 > 9x^2+x+5} \atop {2x-3-x+3\leq5}} \right.\\ \left \{ {{9x^2-27x-36-9x^2-x-5>0} \atop {x\leq5}} \right.\\ \left \{ {{-28x-41>0} \atop {x\leq5}} \right.\\ \left \{ {{-28x>41} \atop {x\leq5}} \right.\\ \left \{ {{x<-\frac{41}{28}} \atop {x\leq5}} \right.\\ x<-\frac{41}{28} " align="absmiddle" class="latex-formula">