Question

х^2+10х+11>0 ; х^2-3х+2≥0 развязать неравенство
(х-2,5)(х+4)≤0 ; х(х+3,2)>0 развязать неравенство аналитичным способом
(х+4)(х+5,6)<0 ; х(х-5)(х+5)≥0 развязать неравенство методом интервалов
помогите срочно ! заранее спасибо )

Answer 1

$x^2+10x+11 textgreater 0, \ x^2+10x+11=0, \ D_{/4}=5^2-11=14, \ x=-5pmsqrt{14}, \ begin{array}{c|ccccc}x&(-infty;-5-sqrt{14})&-5-sqrt{14}&(-5-sqrt{14};-5-sqrt{14})&-5+sqrt{14}&(-5+sqrt{14};+infty)\x^2+10x+11&+&0&-&0&+end{array} \ (x+5+sqrt{14})(x+5-sqrt{14}) textgreater 0, \ xin(-infty;-5-sqrt{14})cup(-5+sqrt{14};+infty).$

$x^2-3x+2geq0; \ x^2-3x+2=0;, \ x_1=1, x_2=2, \ (x-1)(x-2)geq0, \ begin{array}{c|ccccc}x&(-infty;1)&1&(1;2)&2&(2;+infty)\x^2-3x+2&+&0&-&0&+end{array} \ \ (-infty;1]cup[2;+infty)$

$(x-2,5)(x+4)leq0; \ left [ {{ left { {{x-2,5 leq 0,} atop {x+4 geq 0,}} right. } atop { left { {{x-2,5 geq 0,} atop {x+4 leq 0;}} right. }} right. left [ {{ left { {{x leq 2,5,} atop {x geq -4,}} right. } atop { left { {{x geq 2,5,} atop {x leq -4;}} right. }} right. left [ {{-4 leq x leq 2,5,} atop {xinvarnothing;}} right. \ -4 leq x leq 2,5, \ xin[-4;2,5].$

$x(x+3,2) textgreater 0; \ left [ {{ left { {{x textless 0,} atop {x+3,2 textless 0,}} right. } atop { left { {{x textgreater 0,} atop {x+3,2 textgreater 0;}} right. }} right. left [ {{ left { {{x textless 0,} atop {x textless -3,2,}} right. } atop { left { {{x textgreater 0,} atop {x textgreater -3,2;}} right. }} right. left [ {{x textless -3,2,} atop {x textgreater 0;}} right. \xin(-infty;-3,2)cup(0;+infty).$

$(x+4)(x+5,6) textless 0; \ (x+4)(x+5,6)=0, \ x_1=-5,6, x_2=-4; \ begin{array}{c|ccccc}x&(-infty;-5,6)&-5,6&(-5,6;-4)&-4&(-4;+infty)\(x+4)(x+5,6)&+&0&-&0&+end{array} \ \ xin(-5,6;-4).$

$x(x-5)(x+5) geq 0; \ x(x-5)(x+5)=0, \ x_1=-5, x_2=0, x_3=5 \ begin{array}{c|ccccccc}x&(-infty;-5)&-5&(-5;0)&0&(0;5)&5&(5;+infty)\x(x-5)(x+5)&-&0&+&0&-&0&+end{array} \ \ xin[-5;0]cup[5;+infty).$