Решение
log₂(x + y) + log(₂²) (x - y)² = 5
3^(1 + log₃ (x - y)² = 48
ОДЗ: x + y > 0
x - y > 0, x > y
log₂[(x + y)*(x - y)] = 5
3*3^[log₃ (x - y)²] = 48
(x + y)*(x - y)] = 2⁵
(x - y)² = 16
(x - y)² = 4²
x - y = - 4 не удовлетворяет ОДЗ
x - y = 4
(x + y)*( 4)= 32
x + y = 8
y = 8 - x
x - ( 8 - x) = 4
2x = 12
x = 6
y = 8 - 6 = 2
Ответ: (6;2)
Найдём cosα с помощью основного тригонометрического тождества
![\cos^2\alpha + \sin^2\alpha = 1\\\\\cos^2\alpha = 1 - \sin^2\alpha\\\\\cos^2\alpha = 1 - \frac{144}{169}\\\\\cos^2\alpha = \frac{25}{169}\\\\\cos\alpha = \pm\;\frac{5}{13}](https://tex.z-dn.net/?f=%5Ccos%5E2%5Calpha+%2B+%5Csin%5E2%5Calpha+%3D+1%5C%5C%5C%5C%5Ccos%5E2%5Calpha+%3D+1+-+%5Csin%5E2%5Calpha%5C%5C%5C%5C%5Ccos%5E2%5Calpha+%3D+1+-+%5Cfrac%7B144%7D%7B169%7D%5C%5C%5C%5C%5Ccos%5E2%5Calpha+%3D+%5Cfrac%7B25%7D%7B169%7D%5C%5C%5C%5C%5Ccos%5Calpha+%3D+%5Cpm%5C%3B%5Cfrac%7B5%7D%7B13%7D)
Так как α ∈ (π, 3π/2) то cos(α) = -5/13
Найдём tgα
![tg\alpha = \dfrac{\sin\alpha}{cos\alpha}\\\\\\tg\alpha = -\frac{12}{13} : (-\frac{5}{13}) = \frac{12}{5}](https://tex.z-dn.net/?f=tg%5Calpha+%3D+%5Cdfrac%7B%5Csin%5Calpha%7D%7Bcos%5Calpha%7D%5C%5C%5C%5C%5C%5Ctg%5Calpha+%3D+-%5Cfrac%7B12%7D%7B13%7D+%3A+%28-%5Cfrac%7B5%7D%7B13%7D%29+%3D+%5Cfrac%7B12%7D%7B5%7D)
![2x^2+3x+1=x^2-2x-5\\2x^2-x^2+3x+2x+1+5=0\\x^2+5x+6=0\\D=25-4*1*6=1\\x1=\frac{-5+1}{2} =-2\\x2=\frac{-5-1}{2} =-3\\\\](https://tex.z-dn.net/?f=2x%5E2%2B3x%2B1%3Dx%5E2-2x-5%5C%5C2x%5E2-x%5E2%2B3x%2B2x%2B1%2B5%3D0%5C%5Cx%5E2%2B5x%2B6%3D0%5C%5CD%3D25-4%2A1%2A6%3D1%5C%5Cx1%3D%5Cfrac%7B-5%2B1%7D%7B2%7D+%3D-2%5C%5Cx2%3D%5Cfrac%7B-5-1%7D%7B2%7D+%3D-3%5C%5C%5C%5C)
теперь подставляем найденные иксы в любую из функций и находим игрик:
![y(-2)=2*(-2)^2+3(-2)+1=8-6+1=3\\y(-3)=2*(-3)^2+3(-3)+1=18-9+1=10](https://tex.z-dn.net/?f=y%28-2%29%3D2%2A%28-2%29%5E2%2B3%28-2%29%2B1%3D8-6%2B1%3D3%5C%5Cy%28-3%29%3D2%2A%28-3%29%5E2%2B3%28-3%29%2B1%3D18-9%2B1%3D10)
Ответ: точки пересечения (-2;3);(-3;10)
C3−y2c−yc2+y3=3с-2су-2су+3у=3с-4су+3у.