Question

Те которые выделены карандашом

Answer 1

Ответ:

№113

1)$x^{2} +x-30<0\\x^{2} +6x-5x-30<0\\x*(x+6)-5(x+6)<0\\(x+6)*(x-5)<0\\\left \{ {{x+6<0} \atop {x-5>0}} \right. \\\\\left \{ {{x+6<0} \atop {x-5>0}} \right.$$\left \{ {{x<-6} \atop {x>5}} \right. \\\\\left \{ {{x<-6} \atop {x>5}} \right.$

x ∈ ∅

x ∈ 〈 -6, 5〉

4)$-2x^{2} +7x-6<0\\-2x^{2}+4x+3x-6<0\\-2x*(x-2)+3(x-2)<0\\-(x-2)*(2x-3)<0\\(x-2)*(2x-3)>0\\\left \{ {{x-2>0} \atop {2x-3>0}} \right. \\\\\left \{ {{x-2<0} \atop {2x-3<0}} \right.$$\left \{ {{x>2} \atop {x>\frac{3}{2} }} \right. \\\\\left \{ {{x<2} \atop {x<\frac{3}{2} }} \right.$

x ∈ 〈 2, +∞〉

x ∈ 〈 -∞, $\frac{3}{2}$〉

x ∈ 〈 -∞, $\frac{3}{2}$〉 ∪ 〈 2, +∞〉

7) $16x^{2} -8x+1>0\\(4x-1)^{2} >0\\(4x-1)^{2} =0\\x=\frac{1}{4}$

Выражение истинное кроме случая $x=\frac{1}{4}$, $(4x-1)^{2} >0$

x∈ ℝ ∖ { $\frac{1}{4}$ }

№115

$(2x-1)(x+3)\geq 4\\2x^{2} +6x-x-3\geq 4\\2x^{2} +5x-3\geq 4\\2x^{2} +5x-3-4\geq 0\\2x^{2} +7x-2x-7\geq 0\\x(2x+7)-(2x+7)\geq 0\\(2x+7)(x-1)\geq 0\\\left \{ {{2x+7\geq 0} \atop {x-1\geq 0}} \right. \\\\\left \{ {{2x+7\leq 0} \atop {x-1\leq 0}} \right.$

$\left \{ {{x\geq -\frac{7}{2} } \atop {x\geq 1}} \right. \\\\\left \{ {{x\leq -\frac{7}{2} } \atop {x\leq 1}} \right.$

x ∈ [ 1, +∞ 〉

x ∈ 〈 -∞ , $-\frac{7}{2}$ ]

x ∈ 〈 -∞ , $-\frac{7}{2}$ ] ∪ [ 1, +∞ 〉

№116

$2x^{2} +8x\leq 0\\2x*(x+4)\leq 0\\x(x+4)\leq 0\\\left \{ {{x\leq 0} \atop {x+4\geq 0}} \right. \\\\\left \{ {{x\geq 0} \atop {x+4\leq 0}} \right.$$\left \{ {{x\leq 0} \atop {xx\geq -4}} \right. \\\\\left \{ {{x\geq 0} \atop {x\leq -4}} \right.$

x ∈ [ -4, 0 ]

x ∈ ∅

№117

$y=\sqrt{x^{2} -2x-48} \\0=\sqrt{x^{2} -2x-48}\\\sqrt{x^{2} -2x-48}=0\\x^{2} -2x-48=0\\x^{2} +6x-8x-48=0\\x(x+6)-8(x+6)=0\\(x+6)(x-8)=0\\x+6=0\\x-8=0\\x=-6\\x=8\\x1=-6 ,x2=8$