<span>к графику функции f(x)=4x-x^2 проведены касательные в точках с абциссами х1=1 и х2=4. Найдите площадь треугольника образованного этими касательными и осью Ох.</span>
![3a(5ab^3-3)+5a^2b^2(3b-2a)=15a(2ab^3-1)+18](https://tex.z-dn.net/?f=3a%285ab%5E3-3%29%2B5a%5E2b%5E2%283b-2a%29%3D15a%282ab%5E3-1%29%2B18)
Сделаем преобразования
![15a^2b^3-9a+15a^2b^3-10a^3b^2=30a^2b^3-15a+18 \\ \\ 6a - 10a^3b^2=18 \ |2](https://tex.z-dn.net/?f=15a%5E2b%5E3-9a%2B15a%5E2b%5E3-10a%5E3b%5E2%3D30a%5E2b%5E3-15a%2B18+%5C%5C++%5C%5C+6a+-+10a%5E3b%5E2%3D18+%5C+%7C2+)
![3a - 5a^3b^2=9](https://tex.z-dn.net/?f=3a+-+5a%5E3b%5E2%3D9)
Выразим "b" через "а"
![3a - 9 = 5a^3b^2 \\ \\ b^2 = \frac{3a - 9}{5a^3} \ ; \ a \neq 0](https://tex.z-dn.net/?f=3a+-+9+%3D+5a%5E3b%5E2++%5C%5C++%5C%5C+b%5E2+%3D++%5Cfrac%7B3a+-+9%7D%7B5a%5E3%7D+%5C+%3B+%5C+a+%5Cneq+0)
![b = \pm \sqrt{\frac{3a - 9}{5a^3}}](https://tex.z-dn.net/?f=b+%3D+%5Cpm++%5Csqrt%7B%5Cfrac%7B3a+-+9%7D%7B5a%5E3%7D%7D)
ОДЗ:
![\frac{3a - 9}{5a^3} \geq 0](https://tex.z-dn.net/?f=%5Cfrac%7B3a+-+9%7D%7B5a%5E3%7D++%5Cgeq+0)
1)
![\left \{ {{3a - 9 \geq 0} \atop {5a^3\ \textgreater \ 0}} \right. \ \Rightarrow \ \left \{ {{a \geq 3} \atop {a\ \textgreater \ 0}} \right. \ \Rightarrow \ a \geq 3](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7B3a+-+9++%5Cgeq+0%7D+%5Catop+%7B5a%5E3%5C+%5Ctextgreater+%5C+0%7D%7D+%5Cright.++%5C+%09%5CRightarrow+%5C+++%5Cleft+%5C%7B+%7B%7Ba+%5Cgeq+3%7D+%5Catop+%7Ba%5C+%5Ctextgreater+%5C+0%7D%7D+%5Cright.+%5C+%09%5CRightarrow+%5C++a++%5Cgeq+3)
2)
Ответ: ![a \geq 3 \ \bigcup \ a \ \textless \ 0](https://tex.z-dn.net/?f=a++%5Cgeq+3+%5C++%5Cbigcup+%5C+a+%5C+%5Ctextless+%5C+0)
Проверка:
1) ![a \geq 3](https://tex.z-dn.net/?f=a+%5Cgeq+3+)
Пусть а =3, тогда
![b = \pm \sqrt{\frac{3*3 - 9}{5*3^3}} =0](https://tex.z-dn.net/?f=b+%3D+%5Cpm+%5Csqrt%7B%5Cfrac%7B3%2A3+-+9%7D%7B5%2A3%5E3%7D%7D+%3D0)
Подставим "a" и "b" в выражение
![3a - 5a^3b^2=9](https://tex.z-dn.net/?f=3a+-+5a%5E3b%5E2%3D9)
![3 * 3 - 5*3^3*0^2 = 9 \\ \\ 9 = 9](https://tex.z-dn.net/?f=3+%2A+3+-+5%2A3%5E3%2A0%5E2+%3D+9+%5C%5C++%5C%5C+9+%3D+9)
Решение верное!
Пусть а =6, тогда
![b = \pm \sqrt{\frac{3*6 - 9}{5*6^3}} = \pm \sqrt{\frac{9}{5*216}} = \pm \sqrt{\frac{1}{120}}](https://tex.z-dn.net/?f=b+%3D+%5Cpm+%5Csqrt%7B%5Cfrac%7B3%2A6+-+9%7D%7B5%2A6%5E3%7D%7D+%3D+%5Cpm+%5Csqrt%7B%5Cfrac%7B9%7D%7B5%2A216%7D%7D+%3D+%5Cpm+%5Csqrt%7B%5Cfrac%7B1%7D%7B120%7D%7D)
Подставим
![3a - 5a^3b^2= 3 * 6 - 5 *6^3 * (\pm \sqrt{ \frac{1}{120}})^2 = 18 - 9 = 9 \\ \\ 9 = 9](https://tex.z-dn.net/?f=3a+-+5a%5E3b%5E2%3D+3+%2A+6+-+5+%2A6%5E3+%2A+%28%5Cpm+%5Csqrt%7B+%5Cfrac%7B1%7D%7B120%7D%7D%29%5E2+%3D+18+-+9+%3D+9+%5C%5C+%5C%5C+9+%3D+9+)
Решение верное!
2)
![a \ \textless \ 0](https://tex.z-dn.net/?f=a+%5C+%5Ctextless+%5C++0)
Пусть а = -1, тогда
![b = \pm \sqrt{\frac{3*(-1) - 9}{5*(-1)^3}} = \pm \sqrt{\frac{- 12}{-5}} = \pm \sqrt{\frac{12}{5}}](https://tex.z-dn.net/?f=b+%3D+%5Cpm+%5Csqrt%7B%5Cfrac%7B3%2A%28-1%29+-+9%7D%7B5%2A%28-1%29%5E3%7D%7D+%3D+%5Cpm+%5Csqrt%7B%5Cfrac%7B-+12%7D%7B-5%7D%7D+%3D+%5Cpm+%5Csqrt%7B%5Cfrac%7B12%7D%7B5%7D%7D++)
Подставим
![3*(-1) - 5*(-1)^3 * (\pm \sqrt{ \frac{12}{5} })^2 = -3+12 =9 \\ \\ 9 = 9](https://tex.z-dn.net/?f=3%2A%28-1%29+-+5%2A%28-1%29%5E3+%2A+%28%5Cpm+%5Csqrt%7B+%5Cfrac%7B12%7D%7B5%7D+%7D%29%5E2+%3D+-3%2B12+%3D9+%5C%5C+%5C%5C+9+%3D+9)
Решение верное!
A, b, c, d
a; aq; aq^2
aq+d=2aq^2
a+d=32
aq+aq^2=24
d=32-a
aq+d-2aq^2=0
2aq+2aq^2=48
a=16/(3q-1)
2aq(1+q)=48
16q/(3q-1)(1+q)=24
(2q(1+q))/(3q-1)=3
2q^2-7q+3=0
q=3
a=16/(9-1)=2
d=32-2=30
b=2*3=6
c=2*9=18
последовательность 2; 6; 18;30