Помогите с алгебройД, б, г

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Помогите с алгебройД, б, г


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Алгебра (21 баллов) | 22 просмотров
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Правильный ответ

д)

image-7}} \right. \\ \\ \left \{ {{x(x+5)<0} \atop {x>-7}} \right. \\ \\ \left \{ {{x\in(-5;0)} \atop {x\in(-7;+oo)}} \right. \\ \\otvet: x\in(-5;0)\\ \\" alt="\left \{ {{x^2+5x<0} \atop {x>-7}} \right. \\ \\ \left \{ {{x(x+5)<0} \atop {x>-7}} \right. \\ \\ \left \{ {{x\in(-5;0)} \atop {x\in(-7;+oo)}} \right. \\ \\otvet: x\in(-5;0)\\ \\" align="absmiddle" class="latex-formula">

б)

image0}} \right. \\ \\ D=39^2+4*21*6=1521+504=2025=45^2\\ \\ x_{1}= (-39-45)/42=-2\\ \\ x_{2}= (-39+45)/42=1/7\\ \\ \left \{ {{x\in(-2;1/7)} \atop {x\in(0;+oo)}} \right. \\ \\ otvet:x\in(0;1/7)\\ \\" alt="\left \{ {{21x^2+39x-6<0} \atop {x>0}} \right. \\ \\ D=39^2+4*21*6=1521+504=2025=45^2\\ \\ x_{1}= (-39-45)/42=-2\\ \\ x_{2}= (-39+45)/42=1/7\\ \\ \left \{ {{x\in(-2;1/7)} \atop {x\in(0;+oo)}} \right. \\ \\ otvet:x\in(0;1/7)\\ \\" align="absmiddle" class="latex-formula">

г)

image0} \atop {x-3<0}} \right. \\ \\ \left \{ {{(x-12)(x+12)>0} \atop {x<3}} \right. \\ \\ \left \{ {{x\in(-oo;-12)U(12;+oo)} \atop {x\in(-oo;3)}} \right. \\ \\ otvet:{x\in(-oo;-12)\\ \\" alt="\left \{ {{x^2-144>0} \atop {x-3<0}} \right. \\ \\ \left \{ {{(x-12)(x+12)>0} \atop {x<3}} \right. \\ \\ \left \{ {{x\in(-oo;-12)U(12;+oo)} \atop {x\in(-oo;3)}} \right. \\ \\ otvet:{x\in(-oo;-12)\\ \\" align="absmiddle" class="latex-formula">

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