-2x²+9-10=0 /: (-1)
2x²-9+10=0
D=81-4*2*10=81-80=1
9±1 X1=10/4
x½=--------- = X2=2
4
4 (x – 2y) = 3x + 2y – 64x-8y=3x+2y-6
4x-3x=8y+2y-6
x=10y-6
<span>1)√2· √8=√(2·8)=√16=4
2)√2\√18=√2/√(9·2)=√2/(3√2)=1/3
3)√27· √3 =√(27·3)=√81=9
4)√52\√117=√(4·13)/√(13·9)=2·√13/3·√13=2/3
5)√28· √7=√(4·7)·√7=2√7·√7=2·7=14
6)√23\√2300=√23/(√100·23)=√23/(10√23)=1/10=0,1
7)√13· √52=√13·√(4·13)=√13·2√13=13·2=26
8)√12500\√500</span>=√(25·5·100)/√(5·100)=5·10√5/10√5=5
![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]
<span>(3sin²x+tgx)'=3*2sinx cosx + 1/cos²x= </span>6sinx cosx + 1/cos²x=<span>3sin2x + 1/cos²x</span>