<span>(x-5)(x+5)=x^2-25
(6-y)(6+y)=36-y^2
(a-7)(a+7)=a^2-49
(1-c)(1+c)=1-c^2
(m-4)(m+4)=m^2-16
(n-2)(n+2)=n^2-4
(x-9)(9+×)=x^2-81
(11-d)(d +11)=121-d^2
(4+b)(b-4)=b^2-16
(a+12)(12-a)=144-a^2
(c-3)(3+c)=c^2-9
(8-b)(b+8)=64-b^2
(2x-3)(2x+3)=4x^2-9
(4y-7)(4x+7)=
(8a+5)(5-8a)=25-64a^2
(9c-1)(1+9c)=81c^2-1
(10+3d)(3d-10)=9d^2-100
(11x-6)(6+11x)=121x^2-36
(2m-0,7)(2m+0,7)=4m^2-0.49
(1,2-5y)(1,2+5y)=1.44-25y^2
(8n+0,6)(8n-0,6)=64n^2-0.64
(0,9p+2)(0,9p-2)=0.81p^2-4
(1,1d-8)(8+1,1d)=1.21d^2-64
(0,1a+1)(1-0,1a)=1-0.0.1a^2
(0,7x-3y)(0,7x+3y)=0.49x^2-9y^2
(1,3m-4n)(1,3m+4n)=1.69m^2-16n^2
(5a+0,4d)(0,4d-5a)=0.16d^2-25a^2
(2,5p+6q)(6q-2,5p)=36q^2-6.25p^2
(8c-2,1d)(2,1d+8c)=64c^2-4.41d^2
(0,9x+2y)(0,9x-2y)=0.81x^2-4y^2
(1/4a-0,9b)(1/4a+0,9b)=1/16a^2-0.81b^2
(2/5m-1,5n)(2/5m+1,5n)=4/25m^2-2.25n^2
(3/7x-1/2y)(3/7x+1/2y)=9/49x^2-2/4y^2
(4/9c-9/10d)(9/10d+4/9c)=16/81c^2-81/100d^2
(1/3a+6/7b)(6/7b-1/3a)=36/49b^2-1/9a^2
(1/4a-0,9b)(1/4a+0,9b)=1/16a^2-0.81b^2</span>
64а^3+(2-а)^3=(4а+(2-а))(16а^2-4а (2-а)+(2-а)^2)=
=(4а+2-а)(16а^2-8а+4а^2+4-4а+а^2)=
=(3а+2)(21а^2-12а+4)
Ответ показан во вложении(отметь лучший ответ пожалуйста, тебе не сложно, а мне приятно. Будут вопросы пиши.
<span>Найдите производную функции
</span>
![y=e^{4x+5}](https://tex.z-dn.net/?f=y%3De%5E%7B4x%2B5%7D)
<span>
Решение
</span>
![y'=(e^{4x+5})' =e^{4x+5}*(4x+5)'=e^{4x+5}*((4x)'+(5)')=](https://tex.z-dn.net/?f=y%27%3D%28e%5E%7B4x%2B5%7D%29%27+%3De%5E%7B4x%2B5%7D%2A%284x%2B5%29%27%3De%5E%7B4x%2B5%7D%2A%28%284x%29%27%2B%285%29%27%29%3D)
![e^{4x+5}*(4+0)=4e^{4x+5}](https://tex.z-dn.net/?f=e%5E%7B4x%2B5%7D%2A%284%2B0%29%3D4e%5E%7B4x%2B5%7D)
Если функция записана как
![y=e^{4x}+5](https://tex.z-dn.net/?f=y%3De%5E%7B4x%7D%2B5)
![y'=(e^{4x}+5)' =e^{4x}*(4x)'+(5)'=4e^{4x}](https://tex.z-dn.net/?f=y%27%3D%28e%5E%7B4x%7D%2B5%29%27+%3De%5E%7B4x%7D%2A%284x%29%27%2B%285%29%27%3D4e%5E%7B4x%7D)
(2a+c) (a-3c) + a(2c-a) =2a^2-6ac+ac-3c^2+2ac-a^2=a^2-3ac-3c^2