№1.
2x² - 9x + 4 = 0 | ÷ 2
x² - 4,5x + 2 = 0
По теореме обратной теореме Виета:
х1 + х2 = -(-4,5) = 4,5 ; х1 × х2 = 2
Ответ: х1 + х2 = 4,5 ; х1 × х2 = 2
№2.
х1 = 2 ; х2 = 3 => х1 + х2 = 2 + 3 = 5 ; х1 × х2 = 2 × 3 = 6 => искомое квадратное уравнение: х² - 5х + 6 = 0
Ответ: х² - 5х + 6 = 0
№3.
(х² + 7х + 12)/(х + 3) = ((х + 3)(х + 4))/(х + 3) = х + 4
Р.S. х² + 7х + 12 = 0
По теореме обратной теореме Виета:
х1 + х2 = -7 ; х1 × х2 = 12 => х1 = -3 ; х2 = -4 => х² + 7х + 12 = (х + 3)(х + 4)
D=-5.3
a1=-7.7
a7=-7.7-5.3(7-1)
a7=<span>-7.7-5.3*6
а7=-7.7-31.8
а7=-39.5</span>
<span>1)Sin 3x - Cos 3x = 2
По формуле: </span>Sin x - Cos x =
![\sqrt{2}](https://tex.z-dn.net/?f=+%5Csqrt%7B2%7D+)
(sin(x - π/4)) = 0
![\sqrt{2}](https://tex.z-dn.net/?f=+%5Csqrt%7B2%7D+)
(sin(3x - π/4)) = 2
sin(3x - π/4) =
![\frac{2}{ \sqrt{2}}](https://tex.z-dn.net/?f=+%5Cfrac%7B2%7D%7B+%5Csqrt%7B2%7D%7D+)
3x - π/4 = (-1)^n × arcsin
![\frac{2}{ \sqrt{2}}](https://tex.z-dn.net/?f=+%5Cfrac%7B2%7D%7B+%5Csqrt%7B2%7D%7D+)
+ πn (n ∈ Z)
3x = (-1)^n × arcsin
![\sqrt{2}](https://tex.z-dn.net/?f=%5Csqrt%7B2%7D+)
+ π/4 + πn (n ∈ Z)
x =
![\frac{1}{3}](https://tex.z-dn.net/?f=+%5Cfrac%7B1%7D%7B3%7D+)
× (-1)^n × arcsin
![\sqrt{2}](https://tex.z-dn.net/?f=%5Csqrt%7B2%7D+)
+ π/12 + πn/3 (n ∈ Z)
Ответ: x =
![\frac{1}{3}](https://tex.z-dn.net/?f=+%5Cfrac%7B1%7D%7B3%7D+)
× (-1)^n × arcsin
![\sqrt{2}](https://tex.z-dn.net/?f=%5Csqrt%7B2%7D+)
+ π/12 + πn/3 (n ∈ Z)
Наверное составители примера имели ввиду вариант условия b).
![a)\; \; (1+\frac{3x+x}{3+x}):(\frac{1}{x+1}-\frac{x}{1+2x+x})=\frac{3+x+4x}{3+x}:\frac{1+3x-x(x+1)}{(x+1)(1+3x)}=\\\\=\frac{3+5x}{3+x}:\frac{2x-x^2+1}{(x+1)(1+3x)}=\frac{3+5x}{3+x}\cdot \frac{(x+1)(1+3x)}{-(x^2-2x-1)}=-\frac{(3+5x)(x+1)(1+3x)}{(3+x)(x^2-2x-1)}\; ;](https://tex.z-dn.net/?f=a%29%5C%3B+%5C%3B+%281%2B%5Cfrac%7B3x%2Bx%7D%7B3%2Bx%7D%29%3A%28%5Cfrac%7B1%7D%7Bx%2B1%7D-%5Cfrac%7Bx%7D%7B1%2B2x%2Bx%7D%29%3D%5Cfrac%7B3%2Bx%2B4x%7D%7B3%2Bx%7D%3A%5Cfrac%7B1%2B3x-x%28x%2B1%29%7D%7B%28x%2B1%29%281%2B3x%29%7D%3D%5C%5C%5C%5C%3D%5Cfrac%7B3%2B5x%7D%7B3%2Bx%7D%3A%5Cfrac%7B2x-x%5E2%2B1%7D%7B%28x%2B1%29%281%2B3x%29%7D%3D%5Cfrac%7B3%2B5x%7D%7B3%2Bx%7D%5Ccdot+%5Cfrac%7B%28x%2B1%29%281%2B3x%29%7D%7B-%28x%5E2-2x-1%29%7D%3D-%5Cfrac%7B%283%2B5x%29%28x%2B1%29%281%2B3x%29%7D%7B%283%2Bx%29%28x%5E2-2x-1%29%7D%5C%3B+%3B)
![b)\; \; (1+\frac{3x+x^2}{3+x}):(\frac{1}{x+1}-\frac{x}{1+2x+x^2})=(1+\frac{x(3+x)}{3+x}):(\frac{1}{x+1}-\frac{x}{(x+1)^2})=\\\\=(1+x):\frac{x+1-x}{(x+1)^2}=(1+x):\frac{1}{(x+1)^2}=(1+x)\cdot \frac{(x+1)^2}{1}=(x+1)^3](https://tex.z-dn.net/?f=b%29%5C%3B+%5C%3B+%281%2B%5Cfrac%7B3x%2Bx%5E2%7D%7B3%2Bx%7D%29%3A%28%5Cfrac%7B1%7D%7Bx%2B1%7D-%5Cfrac%7Bx%7D%7B1%2B2x%2Bx%5E2%7D%29%3D%281%2B%5Cfrac%7Bx%283%2Bx%29%7D%7B3%2Bx%7D%29%3A%28%5Cfrac%7B1%7D%7Bx%2B1%7D-%5Cfrac%7Bx%7D%7B%28x%2B1%29%5E2%7D%29%3D%5C%5C%5C%5C%3D%281%2Bx%29%3A%5Cfrac%7Bx%2B1-x%7D%7B%28x%2B1%29%5E2%7D%3D%281%2Bx%29%3A%5Cfrac%7B1%7D%7B%28x%2B1%29%5E2%7D%3D%281%2Bx%29%5Ccdot+%5Cfrac%7B%28x%2B1%29%5E2%7D%7B1%7D%3D%28x%2B1%29%5E3)