Question

помогите пж даю 100 баллов​

Answer 1

Ответ:

1)14х+3,5-4х²-х+4х²+12х+9<0

25х<-12,5; х<-0,5; (-беск.;-0,5)

2)8у²-48у+72+15+40у-3у-8у²>2

-11у>-85; у<85/11; у< 7 8/11; (-беск.; 7 8/11)

3)9z²+2z+1/9-9z²+1,5z-6z+1<0

-2,5z<-10/9; z>10/9:25/10; z>10/9•10/25

z>100/225; z>20/45; z>4/9; (4/9;+беск.)

Answer 2

Ответ:

Пользуемся формулами   $(a\pm b)^2=a^2\pm 2ab+b^2$  .

$1)\ \ (3,5-x)(4x+1)+(2x+3)^2 < 0\\\\14x+3,5-4x^2-x+4x^2+12x+9 < 0\\\\25x+12,5 < 0\ \ ,\ \ \ 25x < -12,5\ \ ,\ \ x < -0,5\ \ ,\ \ \underline{\ x\in (-\infty ;-0,5\ )\ }\\\\\\2)\ \ 8(y-3)^2+(5-y)(3+8y) > 2\\\\8(y^2-6y+9)+(15+40y-3y-8y^2) > 2\\\\8y^2-48y+72+15+37y-8y^2 > 2\\\\-11y +87 > 2\ \ ,\ \ \ -11y > -85\ \ ,\ \ y < \dfrac{85}{11}\ \ ,\ \ y < 7\dfrac{8}{11}\ \ ,\ \ \underline{\ y\in (-\infty \, ;\, 7\dfrac{8}{11}\ )\ }$

$3)\ \ \Big(3z+\dfrac{1}{3}\Big)^2-(1,5z+1)(6z-1) < 0\\\\9z^2+2z+\dfrac{1}{9}-(9z^2-1,5z+6z-1) < 0\\\\9z^2+2z+\dfrac{1}{9}-9z^2+1,5z-6z+1 < 0\\\\-2,5z+\dfrac{10}{9} < 0\ \ ,\ \ \ -2,5z < -\dfrac{10}{9}\ \ ,\ \ z > \dfrac{10}{9\cdot 2,5}\ \ ,\ \ z > \dfrac{10\cdot 10}{9\cdot 25}\ ,\ \\\\z > \dfrac{4}{9}\ \ ,\ \ \underline{\ z\in \Big(\ \dfrac{4}{9}\ ;+\infty \, \Big)\ }$