А.) вынесем а за скобку: a(4b + a)
б.) выносим b^3 за скобку: b^3(5 - 3b^2)
в.) вынесем -4ab за скобку: -4ab(3a^3 + b^2 + 2*a^4*b^6)
а.) выносим a+b за скобку: (a+b)(3a+b)
б.) сгруппируем: (xm + 5m)-(xn + 5n). Вынесем из первой скобки m, а из второй n: m(x+5)-n(x+5). Теперь вынесем x+5 за скобки: (x+5)(m-n)
а.)вынесем 3: 3(a^4-4*b^2).воспользуемся формулой "разность квадратов": a^2-b^2 = (a-b)(a+b)
3(a^4-4*b^2) = 3(a^2 - 2b)(a^2 + 2b)
б.) сначала вынесем "-2":-2(x^2 - 6x + 9). Воспользуемся формулой "квадрат разности":
(a-b)^2 = a^2 - 2ab + b^2
-2(x^2 - 6x + 9) = -2(x-3)^2
в.) воспользуемся формулой разность кубов: a^3 - b^3 = (a-b)(a^2 + ab + b^2)
m^3 - 8k^3 = (m-2k)(m^2 + 2mk + 4k^2)
![1)f(x)=3x-\frac{1}{3}x^{3}-x^{2}\\\\f'(x)=3(x)'-\frac{1}{3}(x^{3})'-(x^{2})'=3-\frac{1}{3}*3x^{2}-2x=3-x^{2}-2x\\\\f'(x)<0\Rightarrow -x^{2}-2x+3<0\\\\x^{2}+2x-3>0\\\\(x-1)(x+3)>0](https://tex.z-dn.net/?f=1%29f%28x%29%3D3x-%5Cfrac%7B1%7D%7B3%7Dx%5E%7B3%7D-x%5E%7B2%7D%5C%5C%5C%5Cf%27%28x%29%3D3%28x%29%27-%5Cfrac%7B1%7D%7B3%7D%28x%5E%7B3%7D%29%27-%28x%5E%7B2%7D%29%27%3D3-%5Cfrac%7B1%7D%7B3%7D%2A3x%5E%7B2%7D-2x%3D3-x%5E%7B2%7D-2x%5C%5C%5C%5Cf%27%28x%29%3C0%5CRightarrow%20-x%5E%7B2%7D-2x%2B3%3C0%5C%5C%5C%5Cx%5E%7B2%7D%2B2x-3%3E0%5C%5C%5C%5C%28x-1%29%28x%2B3%29%3E0)
+ - +
________₀_________₀_________
- 3 1
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Ответ : x ∈ (- ∞ ; - 3) ∪ (1 ; + ∞)
![2)f(x)=\frac{2x}{(x-1)^{2} }\\\\f'(x)=\frac{2(x)'*(x-1)^{2}-2x*((x-1)^{2})'}{(x-1)^{4}}=\frac{2*(x^{2}-2x+1)-2x*2(x-1)}{(x-1)^{4}}=\frac{2x^{2}-4x+2-4x^{2}+4x}{(x-1)^{4}}=\frac{-2x^{2}+2 }{(x-1)^{4}}=\frac{-2(x^{2}-1)}{(x-1)x^{4}}\\\\f'(x)=0\Rightarrow\\\\\left \{ {{x^{2}-1=0 } \atop {x-1\neq0 }} \right. \\\\\\\left \{ {{(x-1)(x+1)=0} \atop {x\neq1 }} \right. \\\\\\\left \{ {{\left[\begin{array}{ccc}x_{1}=-1 \\x_{2} =1-neyd\end{array}\right } \atop {x\neq1 }} \right. \\\\](https://tex.z-dn.net/?f=2%29f%28x%29%3D%5Cfrac%7B2x%7D%7B%28x-1%29%5E%7B2%7D%20%7D%5C%5C%5C%5Cf%27%28x%29%3D%5Cfrac%7B2%28x%29%27%2A%28x-1%29%5E%7B2%7D-2x%2A%28%28x-1%29%5E%7B2%7D%29%27%7D%7B%28x-1%29%5E%7B4%7D%7D%3D%5Cfrac%7B2%2A%28x%5E%7B2%7D-2x%2B1%29-2x%2A2%28x-1%29%7D%7B%28x-1%29%5E%7B4%7D%7D%3D%5Cfrac%7B2x%5E%7B2%7D-4x%2B2-4x%5E%7B2%7D%2B4x%7D%7B%28x-1%29%5E%7B4%7D%7D%3D%5Cfrac%7B-2x%5E%7B2%7D%2B2%20%7D%7B%28x-1%29%5E%7B4%7D%7D%3D%5Cfrac%7B-2%28x%5E%7B2%7D-1%29%7D%7B%28x-1%29x%5E%7B4%7D%7D%5C%5C%5C%5Cf%27%28x%29%3D0%5CRightarrow%5C%5C%5C%5C%5Cleft%20%5C%7B%20%7B%7Bx%5E%7B2%7D-1%3D0%20%7D%20%5Catop%20%7Bx-1%5Cneq0%20%7D%7D%20%5Cright.%20%5C%5C%5C%5C%5C%5C%5Cleft%20%5C%7B%20%7B%7B%28x-1%29%28x%2B1%29%3D0%7D%20%5Catop%20%7Bx%5Cneq1%20%7D%7D%20%5Cright.%20%5C%5C%5C%5C%5C%5C%5Cleft%20%5C%7B%20%7B%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx_%7B1%7D%3D-1%20%5C%5Cx_%7B2%7D%20%3D1-neyd%5Cend%7Barray%7D%5Cright%20%7D%20%5Catop%20%7Bx%5Cneq1%20%7D%7D%20%5Cright.%20%5C%5C%5C%5C)
![Otvet:\boxed{-1}](https://tex.z-dn.net/?f=Otvet%3A%5Cboxed%7B-1%7D)
![3)f(x)=x*Cos2x\\\\f'(x)=(x)'*Cos2x+x*(Cos2x)'=1*Cos2x+x*(-2Sin2x)=Cos2x-2xSin2x\\\\f'(0)=Cos0-2*0*Sin0=1-0=1](https://tex.z-dn.net/?f=3%29f%28x%29%3Dx%2ACos2x%5C%5C%5C%5Cf%27%28x%29%3D%28x%29%27%2ACos2x%2Bx%2A%28Cos2x%29%27%3D1%2ACos2x%2Bx%2A%28-2Sin2x%29%3DCos2x-2xSin2x%5C%5C%5C%5Cf%27%280%29%3DCos0-2%2A0%2ASin0%3D1-0%3D1)
1) 2x=18-x
2x+x=18
3x=18
x= 18:3
x=6
2)7x+3=30-2x
7x+2x=30-3
9x=27
x=27:9
x=3
3)7-2x=3x-18
-2x-3x=-18-7
-5x=-25 умножаем на -1
5x=25
x=25:5
x=5
правильно но это ответ не в стандартном виде
правильно будет 8,3*10^(-7)
как приводить в стандартный вид во вложении
Используем два свойства логарифма
1)
![log_{6}(5) = \frac{1}{ log_{5}(6) }](https://tex.z-dn.net/?f=+log_%7B6%7D%285%29++%3D++%5Cfrac%7B1%7D%7B+log_%7B5%7D%286%29+%7D+)
2)
![\frac{ log_{5}(2) }{ log_{5}(6) } = log_{6}(2)](https://tex.z-dn.net/?f=+%5Cfrac%7B+log_%7B5%7D%282%29+%7D%7B+log_%7B5%7D%286%29+%7D++%3D++log_%7B6%7D%282%29+)
теперь решаем
![3 \times ( log_{6}(2) + log_{6}(3)) = 3 \times log_{6}(6) = 3](https://tex.z-dn.net/?f=3+%5Ctimes+%28+log_%7B6%7D%282%29++%2B++log_%7B6%7D%283%29%29++%3D+3+%5Ctimes++log_%7B6%7D%286%29++%3D+3)
Ответ правильный : 3