1)Найдите значения х,при которых трехчлен -16x^2+8x-1 принимает отрицательные значения.
-16x^2+8x-1=-(16x^2-8x+1)=-(4x-1)^2
(4x-1)^2 - всегда дает положительное значение кроме x=1/4
- (4x-1)^2 - всегда дает отрицателное значение кроме x=1/4
ответ x (-∞;1/4) U (1/4; +∞)
2)докажите,что при любом значении а верно неравенство:
6а<а^2+10
0<а^2-6a+10
0<а^2-6a+9+1
0<а^2-6a+3^2+1
0<(а-3)^2+1
(а-3)^2 - положительное при любом a
значит (а-3)^2+1 - положительное при любом а
значит при любом значении а верно ИСХОДНОЕ неравенство
ДОКАЗАНО
My name is Karoyan. I’m 15 years old. I live in Moskow. It’s one of the biggest and most beautiful cities in Russia. I’m in my last year of school and I’m thinking of entering the Polytechnic University. My parents are always helping me in every possible way with studies and pocket money, but I understand that it’s not easy to earn money. So, this year I’ve decided to find a part-time job to be able to pay for my future studies. I was lucky and I found a job at local McDonalds. It’s a suitable job for me, as I can choose the days and hours when I can work. I chose to work at the second part of the day, as I need to go to school every morning from 8 am till 2 pm. Afterwards I’m free to do what I want. So I go to work which is from 3 pm till 9. I know it’s long hours and I come back home really exhausted but the pay is good. Besides, I don’t work every day. I work only on Tuesdays, Thursdays and Saturdays. I sometimes find it difficult to keep up with my homework but I’m doing fine. I always listen attentively the new topics at the lessons and it helps me to be prepared for the following lessons. My work at McDonalds isn’t easy, but it teaches me how to cooperate with colleagues, how to deal with customers and else. I’m sure it will help me in my future life. Moreover, I always have my own pocket money and I even can buy something for our house. My parents are very proud of me.
ответ:a) y=x^4 -3x^8 +9 y' = 4x^3 - 24x^7
б) y=1/x -16√x y' = -(1/(x^2)) - (8/sqrt(x))
в) y=-3/x -7tgx + x/8 y' = 3/(x^2) - 7/(cos^2(x)) + 1/8
г) y=cosx + 4√x y' = -sinx + 2/sqrt(x)
д) y= 2cosx + 4√x y' = -2sinx + 2/sqrt(x)
а) y=x *ctgx y' = ctgx - (x/(sin^2(x)))
б) y=√x *tgx y' = tgx/2*sqrt(x) + sqrt(x)/cos^2(x)
в) y=sinx/x y' = (cosx*x - sinx) / sin^2(x)
г) y=3x+3/3x-3 = y' = ( (3x+3)' * (3x-3) - (3x+3) * (3x-3)' ) / ((3x-3)^2) = (3(3x-3) - 3(3x+3))/ (3x-3)^2
а) y=(3x-4)^6 y' = 6(3x-4)^5 * 3 = 18(3x-4)^5
б) y=√7x-√3 y' = √7√x -√3 = (√7)/2x + 0 + 0 = (√7)/2x
в) y=sin(3x- π/4)
(c*f(x))' = c*f(x)' умножим потом на -1.
y' = (cos(3x + π/4))' = (cos(3x + π/4))'(3x+π/4)' = -3sin(3x+π/4)
Обратно умножим на -1
3sin(3x+π/4)
Объяснение: там должен быть ответ на твой вопрос удачи!