Помогите,с алгеброй,пожалуйста....2.72(б,в,ж),2.70(в,е) и номер..(г)

Question 1

Помогите,с алгеброй,пожалуйста....
2.72(б,в,ж),2.70(в,е) и номер..(г)

Question 2

$(x^2-1)(x^2-5x+4)<0 \\\ (x-1)(x+1)(x-1)(x-4)<0 \\\ (x+1)(x-4)<0 \\\ x\in(-1;4)$

0 \\\ (x-8)(x+1)(x+2)(x+1)>0 \\\ (x-8)(x+2)>0 \\\ x\in(-\infty; -2)\cup(8; +\infty)" alt="(x^2-7x-8)(x^2+3x+2)>0 \\\ (x-8)(x+1)(x+2)(x+1)>0 \\\ (x-8)(x+2)>0 \\\ x\in(-\infty; -2)\cup(8; +\infty)" align="absmiddle" class="latex-formula">

0 \\\ (x+2)(x+3)(x^2+4)>0 \\\ (x+2)(x+3)>0 \\\ x\in(-\infty; -3)\cup(-2; +\infty)" alt="x^4+5x^3+10x^2+20x+24>0 \\\ (x+2)(x+3)(x^2+4)>0 \\\ (x+2)(x+3)>0 \\\ x\in(-\infty; -3)\cup(-2; +\infty)" align="absmiddle" class="latex-formula">

$(x+5)^2(x+3)(x-3)<0 \\\ (x+3)(x-3)<0 \\\ x\in(-3;3)$

$(x^2-9)(x+2)^2<0 \\\ (x-3)(x+3)<0 \\\ x\in(-3; 3)$

$\left \{ {x^2+2xy-3y^2=0} \atop {y^2+x+y=1.25}} \right. \\\ \left \{ {x^2+2xy+y^2-4y^2=0} \atop {y^2+x+y=1.25}} \right. \\\ \left \{ {(x+y)^2-4y^2=0} \atop {y^2+x+y=1.25}} \right. \\\ \left \{ {(x+y-2y)(x+y+2y)=0} \atop {y^2+x+y=1.25}} \right. \\\ \left \{ {(x-y)(x+3y)=0} \atop {y^2+x+y=1.25}} \right.$

$x-y=0 \\\ y^2+y+y=1.25 \\\ y^2+2y-1.25=0 \\\ D_1=1+1.25=2.25 \\\ y_1=x_1=-2.5 \\\ y_2=x_2=0.5$

$x+3y=0 \\\ y^2-3y+y=1.25 \\\ y^2-2y-1.25=0 \\\ D_1=1+1.25=2.25 \\\ y_1=-0.5 \\\ y_2=2.5 \\\ x_1=1.5 \\\ x_2=-7.5$
Ответ: (-2,5; -2,5); (0,5; 0,5); (1,5; -0,5); (-7,5; 2,5)

Artem112_zn · Answer 1 · 2018-03-14T16:09:39+0000

$(x^2-1)(x^2-5x+4)<0 \\\ (x-1)(x+1)(x-1)(x-4)<0 \\\ (x+1)(x-4)<0 \\\ x\in(-1;4)$

0 \\\ (x-8)(x+1)(x+2)(x+1)>0 \\\ (x-8)(x+2)>0 \\\ x\in(-\infty; -2)\cup(8; +\infty)" alt="(x^2-7x-8)(x^2+3x+2)>0 \\\ (x-8)(x+1)(x+2)(x+1)>0 \\\ (x-8)(x+2)>0 \\\ x\in(-\infty; -2)\cup(8; +\infty)" align="absmiddle" class="latex-formula">

0 \\\ (x+2)(x+3)(x^2+4)>0 \\\ (x+2)(x+3)>0 \\\ x\in(-\infty; -3)\cup(-2; +\infty)" alt="x^4+5x^3+10x^2+20x+24>0 \\\ (x+2)(x+3)(x^2+4)>0 \\\ (x+2)(x+3)>0 \\\ x\in(-\infty; -3)\cup(-2; +\infty)" align="absmiddle" class="latex-formula">

$(x+5)^2(x+3)(x-3)<0 \\\ (x+3)(x-3)<0 \\\ x\in(-3;3)$

$(x^2-9)(x+2)^2<0 \\\ (x-3)(x+3)<0 \\\ x\in(-3; 3)$

$\left \{ {x^2+2xy-3y^2=0} \atop {y^2+x+y=1.25}} \right. \\\ \left \{ {x^2+2xy+y^2-4y^2=0} \atop {y^2+x+y=1.25}} \right. \\\ \left \{ {(x+y)^2-4y^2=0} \atop {y^2+x+y=1.25}} \right. \\\ \left \{ {(x+y-2y)(x+y+2y)=0} \atop {y^2+x+y=1.25}} \right. \\\ \left \{ {(x-y)(x+3y)=0} \atop {y^2+x+y=1.25}} \right.$

$x-y=0 \\\ y^2+y+y=1.25 \\\ y^2+2y-1.25=0 \\\ D_1=1+1.25=2.25 \\\ y_1=x_1=-2.5 \\\ y_2=x_2=0.5$

$x+3y=0 \\\ y^2-3y+y=1.25 \\\ y^2-2y-1.25=0 \\\ D_1=1+1.25=2.25 \\\ y_1=-0.5 \\\ y_2=2.5 \\\ x_1=1.5 \\\ x_2=-7.5$
Ответ: (-2,5; -2,5); (0,5; 0,5); (1,5; -0,5); (-7,5; 2,5)