Выразим у через х
у=(2-3х) /4
х=1/3, у=1/4
х=1/6, у=15/40 =3/8
<span>х=0,1, у=17/40 </span>
![( \frac{5}{3})^{ \frac{ x^{2}+x-3}{x+1}} \leq \frac{2}{3}* (\frac{5}{2})^{ \frac{ x^{2}+x-3}{x+1}}](https://tex.z-dn.net/?f=%28+%5Cfrac%7B5%7D%7B3%7D%29%5E%7B+%5Cfrac%7B+x%5E%7B2%7D%2Bx-3%7D%7Bx%2B1%7D%7D+%5Cleq++%5Cfrac%7B2%7D%7B3%7D%2A+%28%5Cfrac%7B5%7D%7B2%7D%29%5E%7B+%5Cfrac%7B+x%5E%7B2%7D%2Bx-3%7D%7Bx%2B1%7D%7D+++++)
ОДЗ: x≠-1
Пусть
![{ \frac{ x^{2}+x-3}{x+1}} =t](https://tex.z-dn.net/?f=%7B+%5Cfrac%7B+x%5E%7B2%7D%2Bx-3%7D%7Bx%2B1%7D%7D+%3Dt)
, тогда получаем
![(\frac{5}{3}^{t}) \leq \frac{2}{3}* (\frac{5}{2})^{t}](https://tex.z-dn.net/?f=+%28%5Cfrac%7B5%7D%7B3%7D%5E%7Bt%7D%29+%5Cleq++%5Cfrac%7B2%7D%7B3%7D%2A+%28%5Cfrac%7B5%7D%7B2%7D%29%5E%7Bt%7D++++)
![( \frac{5}{3}* \frac{2}{5})^t \leq \frac{2}{3}](https://tex.z-dn.net/?f=%28+%5Cfrac%7B5%7D%7B3%7D%2A+%5Cfrac%7B2%7D%7B5%7D%29%5Et+%5Cleq++%5Cfrac%7B2%7D%7B3%7D+++)
⇒
![( \frac{2}{3})^t \leq (\frac{2}{3})^1](https://tex.z-dn.net/?f=%28+%5Cfrac%7B2%7D%7B3%7D%29%5Et+%5Cleq+%28%5Cfrac%7B2%7D%7B3%7D%29%5E1+)
⇒ t≥1
![{ \frac{ x^{2}+x-3}{x+1}} \geq 1](https://tex.z-dn.net/?f=%7B+%5Cfrac%7B+x%5E%7B2%7D%2Bx-3%7D%7Bx%2B1%7D%7D++%5Cgeq+1)
⇒
![{ \frac{ x^{2}+x-3-x-1}{x+1}} \geq 0](https://tex.z-dn.net/?f=%7B+%5Cfrac%7B+x%5E%7B2%7D%2Bx-3-x-1%7D%7Bx%2B1%7D%7D++%5Cgeq+0)
⇒
![{ \frac{ x^{2}-4}{x+1}} \geq 0](https://tex.z-dn.net/?f=%7B+%5Cfrac%7B+x%5E%7B2%7D-4%7D%7Bx%2B1%7D%7D++%5Cgeq+0)
![{ \frac{ x^{2}-4}{x+1}} =0](https://tex.z-dn.net/?f=%7B+%5Cfrac%7B+x%5E%7B2%7D-4%7D%7Bx%2B1%7D%7D+%3D0)
x1=-2; x2=2; x3=-1
- + - +
-----------●---------0---------<span>●--------->
-2 -1 2 x
x</span>∈[-2;-1)∪[2; +∞]
У'=7-6cosx
7-6cosx=0
-6cosx=-7
Cosx=7/6
нет корней, т.к. -1Y(-p/2)=-7p/2+6p/2+8
y(0)=0-0+8=8
Ответ: 8