![log_{0,8} \frac{(2x-4)}{(8 - x)} \geq 0 \\ \\ 2x - 4 \ \textgreater \ 0 \\ x\ \textgreater \ 2 \\ 8 - x \ \textgreater \ 0 \\ x\ \textless \ 8 \\ log_{0,8}(2x-4) \geq log_{0,8}(8-x) \\ 2x - 4 \leq 8-x \\ 3x \leq 12 ](https://tex.z-dn.net/?f=log_%7B0%2C8%7D+%5Cfrac%7B%282x-4%29%7D%7B%288+-+x%29%7D++%5Cgeq++0+%5C%5C++%5C%5C+%0A2x+-+4+%5C+%5Ctextgreater+%5C++0+%5C%5C+%0Ax%5C+%5Ctextgreater+%5C+2+%5C%5C+8+-+x+%5C+%5Ctextgreater+%5C++0+%5C%5C+x%5C+%5Ctextless+%5C+8+%5C%5C+log_%7B0%2C8%7D%282x-4%29++%5Cgeq+log_%7B0%2C8%7D%288-x%29+%5C%5C+2x+-+4+%5Cleq+8-x+%5C%5C+3x+%5Cleq+12%0A++)
Учитывая ОДЗ, находим промежуток ответов: (2; 4]
Правильные ответы А (3) и В (-0,5)
<span>3а (а – b) + (b (2a – b)=3a</span>²-3ab+2ab-b²=3a²-ab+b²<span>
3с (с – 2) – (с – 3)</span>²=3c²-6c-c²+6c-9=2c²-9<span>
(3х + 1) (4х – 2) – 6 (2х – 1)</span>²<span>+ 14=12x</span>² + 4x - 6x - 2 - 24x² + 24x - 6=-12x²+22x-8
(3g-20g^2)/5=0
g(3-20g)/5=0
g1=0
3-20g=0
20g=3
g=3/20