Question

Помогите пожалуйста
289.Найдите сумму целых решений неравенств​

Answer 1

Ответ: $-1+3+2+3+(-1)+(-1) = -1+3+2+3-1-1=5$

Объяснение:

а) $\frac{x^2-2x-3}{-2} > 0$    

$x^2-3x-3<0$

$x^2+x-3x-3<0$

$x*(x+1)-3(x+1)<0$

$(x+1)*(x-3)<0$

$\left \{ {{x+1<0} \atop {x-3>0}} \right.$

$\left \{ {{x+1>0} \atop {x-3<0}} \right.$

$\left \{ {{x<-1} \atop {x>3}} \right.$

$\left \{ {{x>-1} \atop {x<3}} \right.$

x∈∅

x∈ ( -1; 3 )

b) $\frac{x^2-5x+6}{-4} \geq 0$

$x^2-5x+6\leq 0$

$x^2-2x-3x+6\leq 0$

$x*(x-2)-3(x-2)\leq 0$

$(x-2)*(x-3)\leq 0$

$\left \{ {{x-2\leq 0} \atop {x-3\geq 0}} \right.$

$\left \{ {{x-2\geq0 } \atop {x-3\leq0 }} \right.$

$\left \{ {{x\leq 2 } \atop {x\geq 3 }} \right.$

$\left \{ {{x\geq 2 } \atop {x\leq 3 }} \right.$

x∈∅

x∈ [ 2; 3 ]

с) $\frac{2x^2+x-1}{\pi-4 } \geq 0$

$2x^2+x-1\leq 0$

$2x^2+2x-x-1\leq 0$

$2x*(x+1)-(x+1)\leq 0$

$(x+1)*(2x-1)\leq 0$

$\left \{ {{x+1\leq 0} \atop {2x-1\geq0 }} \right.$

$\left \{ {{x+1\geq0 } \atop {2x-1\leq0 }} \right.$

$\left \{ {{x\leq -1 } \atop {x\geq \frac{1}{2} }} \right.$

$\left \{ {{x\geq-1 } \atop {x\leq \frac{1}{2} }} \right.$

x∈∅

x∈ [ -1; $\frac{1}{2}$ ]

d) $\frac{3x^2+5x+2}{\sqrt{3}-5 } >0$

$3x^2+5x+2<0$

$3x^2+3x+2x+2<0$

$3x*(x+1)+2(x+1)<0$

$(x+1)*(3x+2)<0$

$\left \{ {{x+1<0} \atop {3x+2>0}} \right.$

$\left \{ {{x+1>0} \atop {3x+2<0}} \right.$

$\left \{ {{x<-1} \atop {x>-\frac{2}{3 }} \right.$

$\left \{ {{x>-1} \atop {x<-\frac{2}{3} }} \right.$

x∈∅

x∈ ( -1 ; - $\frac{2}{3}$ )