A)4-16aˇ2 = 4.(1-4aˇ2)=4.(1+2a)(1-2a)
b)9bˇ3 -bˇ3 = 8bˇ3
3xˇ3 +6xˇ2=12x+24
3xˇ2(x+2) = 12(x+2)
3xˇ2(x+2) -12(x+2)=0
xˇ2(x+2) - 4(x+2)=0
(x+2)(xˇ2-4)=0
(x+2)(x+2)(x-2)=0
a)x+2 =0 , x=-2
b)x+2=0, x=-2
c)x-2=0, x=2
производная y=-8x^7+8X^3, приравниваешь к нулю, получается х=0 или х=+-1, точки максимума -1 и +1, подставляешь в исходное уравнение, получается 2
![log_{0,5} \frac{4+x}{x-1} \geq 2 2= log_{0,5} 0,5^{2} = log_{0,5} 0,25](https://tex.z-dn.net/?f=+log_%7B0%2C5%7D++%5Cfrac%7B4%2Bx%7D%7Bx-1%7D++%5Cgeq+2%0A%0A2%3D+log_%7B0%2C5%7D+0%2C5%5E%7B2%7D++%3D+log_%7B0%2C5%7D+0%2C25)
![log_{0,5} \frac{4+x}{x-1} \geq log_{0,5} 0,25](https://tex.z-dn.net/?f=+log_%7B0%2C5%7D++%5Cfrac%7B4%2Bx%7D%7Bx-1%7D++%5Cgeq++log_%7B0%2C5%7D+0%2C25)
основание логарифма
а=0,5. 0<0,5<1
знак неравенства меняем:
![\left \{ {{ \frac{4+x}{x-1} \ \textgreater \ 0} \atop { \frac{4+x}{x-1} \leq 0,25 }} \right. \left \{ {{ \frac{4+x}{x-1} \ \textgreater \ 0} \atop { \frac{4+x}{x-1}-0,25 \leq 0 }} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7B+%5Cfrac%7B4%2Bx%7D%7Bx-1%7D++%5C+%5Ctextgreater+%5C+0%7D+%5Catop+%7B+%5Cfrac%7B4%2Bx%7D%7Bx-1%7D+%5Cleq+0%2C25+%7D%7D+%5Cright.++++%5Cleft+%5C%7B+%7B%7B+%5Cfrac%7B4%2Bx%7D%7Bx-1%7D+%5C+%5Ctextgreater+%5C+0%7D+%5Catop+%7B+%5Cfrac%7B4%2Bx%7D%7Bx-1%7D-0%2C25+%5Cleq+0+%7D%7D+%5Cright.+)
![1. \frac{4+x}{x-1} \ \textgreater \ 0](https://tex.z-dn.net/?f=+1.++++%5Cfrac%7B4%2Bx%7D%7Bx-1%7D+%5C+%5Ctextgreater+%5C+0)
,метод интервалов:
![\left \{ {{x+4=0} \atop {x-1 \neq 0}} \right. \left \{ {{x=-4} \atop {x \neq 1}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bx%2B4%3D0%7D+%5Catop+%7Bx-1+%5Cneq+0%7D%7D+%5Cright.+++++++%5Cleft+%5C%7B+%7B%7Bx%3D-4%7D+%5Catop+%7Bx+%5Cneq+1%7D%7D+%5Cright.+)
+ - +
-----------(-4)-----------(1)----------------->x
x∈(-4;∞)∪(1;∞)
![2. \frac{4+x}{x-1}-0,25 \leq 0.](https://tex.z-dn.net/?f=2.++++++++%5Cfrac%7B4%2Bx%7D%7Bx-1%7D-0%2C25+%5Cleq+0.)
![\frac{4+x-0,25x+0,25}{x-1} \leq 0, \frac{0,75x+4,25}{x-1} \leq 0](https://tex.z-dn.net/?f=+%5Cfrac%7B4%2Bx-0%2C25x%2B0%2C25%7D%7Bx-1%7D++%5Cleq+0%2C+++++++%5Cfrac%7B0%2C75x%2B4%2C25%7D%7Bx-1%7D++%5Cleq+0)
метод интервалов:
![\left \{ {{0,75x+4,25=0} \atop {x-1 \neq 0}} \right. \left \{ {{x= -\frac{17}{3} } \atop {x \neq 1}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7B0%2C75x%2B4%2C25%3D0%7D+%5Catop+%7Bx-1+%5Cneq+0%7D%7D+%5Cright.++++++%5Cleft+%5C%7B+%7B%7Bx%3D+-%5Cfrac%7B17%7D%7B3%7D+%7D+%5Catop+%7Bx+%5Cneq+1%7D%7D+%5Cright.+)
+ - +
----------[-17/3]-------------(1)--------------->x
x∈[-17/3;1)
/ / / / / / / / / / / / / / / /
------------[-17/3]------(-4)-----------(1)------------------>x
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
x∈[-17/3;-4)
(а-в)²=(в-а)²
значит ответ 0