1-2(3x+4)=5+6x
1-6x-8=5+6x
1-8-5=12x
-12=12x
x=-1
х(квадрат)-3х-10,Д=9+40=49, х1=3+7/2=5,х2=3-7/2=-2
а) ![x^{2}-9=0](https://tex.z-dn.net/?f=x%5E%7B2%7D-9%3D0)
![x^{2}=9](https://tex.z-dn.net/?f=x%5E%7B2%7D%3D9)
![x=\pm\sqrt{9}](https://tex.z-dn.net/?f=x%3D%5Cpm%5Csqrt%7B9%7D)
![x_{1}=-3](https://tex.z-dn.net/?f=x_%7B1%7D%3D-3)
![x_{2}=3](https://tex.z-dn.net/?f=x_%7B2%7D%3D3)
Ответ:
; ![x_{2}=3](https://tex.z-dn.net/?f=x_%7B2%7D%3D3)
==========================================
б) ![x^{2}+4x=0](https://tex.z-dn.net/?f=x%5E%7B2%7D%2B4x%3D0)
![x(x+4)=0](https://tex.z-dn.net/?f=x%28x%2B4%29%3D0)
![x_{1}=0](https://tex.z-dn.net/?f=x_%7B1%7D%3D0)
![x+4=0](https://tex.z-dn.net/?f=x%2B4%3D0)
![x_{2}=-4](https://tex.z-dn.net/?f=x_%7B2%7D%3D-4)
Ответ:
; ![x_{2}=-4](https://tex.z-dn.net/?f=x_%7B2%7D%3D-4)
=========================================
в) ![x^{2}+10=0](https://tex.z-dn.net/?f=x%5E%7B2%7D%2B10%3D0)
![x^{2}=-10](https://tex.z-dn.net/?f=x%5E%7B2%7D%3D-10)
Ответ: корней нет.
=========================================
г) ![x^{2}+5x-6=0](https://tex.z-dn.net/?f=x%5E%7B2%7D%2B5x-6%3D0)
<span>Cчитаем дискриминант:</span>
![D=5^{2}-4\cdot1\cdot(-6)=25+24=49](https://tex.z-dn.net/?f=D%3D5%5E%7B2%7D-4%5Ccdot1%5Ccdot%28-6%29%3D25%2B24%3D49)
<span>Дискриминант положительный</span>
![\sqrt{D}=7](https://tex.z-dn.net/?f=%5Csqrt%7BD%7D%3D7)
<span>Уравнение имеет два различных корня:</span>
![x_{1}=\frac{-5+7}{2\cdot1}=\frac{2}{2}=1](https://tex.z-dn.net/?f=x_%7B1%7D%3D%5Cfrac%7B-5%2B7%7D%7B2%5Ccdot1%7D%3D%5Cfrac%7B2%7D%7B2%7D%3D1)
![x_{2}=\frac{-5-7}{2\cdot1}=\frac{-12}{2}=-6](https://tex.z-dn.net/?f=x_%7B2%7D%3D%5Cfrac%7B-5-7%7D%7B2%5Ccdot1%7D%3D%5Cfrac%7B-12%7D%7B2%7D%3D-6)
Ответ:
; ![x_{2}=-6](https://tex.z-dn.net/?f=x_%7B2%7D%3D-6)
=========================================
д) ![3x^{2}-5x-8=0](https://tex.z-dn.net/?f=3x%5E%7B2%7D-5x-8%3D0)
<span>Cчитаем дискриминант:</span>
![D=(-5)^{2}-4\cdot3\cdot(-8)=25+96=121](https://tex.z-dn.net/?f=D%3D%28-5%29%5E%7B2%7D-4%5Ccdot3%5Ccdot%28-8%29%3D25%2B96%3D121)
<span>Дискриминант положительный</span>
![\sqrt{D}=11](https://tex.z-dn.net/?f=%5Csqrt%7BD%7D%3D11)
<span>Уравнение имеет два различных корня:</span>
![x_{1}=\frac{5+11}{2\cdot3}=\frac{16}{6}=\frac{8}{3}=2\frac{2}{3}](https://tex.z-dn.net/?f=x_%7B1%7D%3D%5Cfrac%7B5%2B11%7D%7B2%5Ccdot3%7D%3D%5Cfrac%7B16%7D%7B6%7D%3D%5Cfrac%7B8%7D%7B3%7D%3D2%5Cfrac%7B2%7D%7B3%7D)
![x_{2}=\frac{5-11}{2\cdot3}=\frac{-6}{6}=-1](https://tex.z-dn.net/?f=x_%7B2%7D%3D%5Cfrac%7B5-11%7D%7B2%5Ccdot3%7D%3D%5Cfrac%7B-6%7D%7B6%7D%3D-1)
Ответ:
; ![x_{2}=-1](https://tex.z-dn.net/?f=x_%7B2%7D%3D-1)
С помощью бинома Ньютона:
![(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n - k} b^k](https://tex.z-dn.net/?f=%28a%2Bb%29%5En+%3D+%5Csum_%7Bk%3D0%7D%5En+%5Cbinom%7Bn%7D%7Bk%7D+a%5E%7Bn+-+k%7D+b%5Ek+)
В нашем случае:
![(x+1)^{50}=\sum_{k=0}^{50} \binom{50}{k}x^{50 - k}](https://tex.z-dn.net/?f=%28x%2B1%29%5E%7B50%7D%3D%5Csum_%7Bk%3D0%7D%5E%7B50%7D+%5Cbinom%7B50%7D%7Bk%7Dx%5E%7B50+-+k%7D)
(здесь не вместилось, поэтому все решение в картинке :D)
P.S
Не пугайтесь :D