Решение задания приложено
Ответ: (2x-1+3y)(2x-1-3y)
![\mathop{\mathrm{tg}}\left(\alpha+\dfrac\pi4\right)=\dfrac{\mathop{\mathrm{tg}}\alpha+\mathop{\mathrm{tg}}\frac\pi4}{1-\mathop{\mathrm{tg}}\alpha\mathop{\mathrm{tg}}\frac\pi4}=\dfrac{1+\mathop{\mathrm{tg}}\alpha}{1-\mathop{\mathrm{tg}}\alpha}=-\sqrt2\\ 1+\mathop{\mathrm{tg}}\alpha=\sqrt{2}(\mathop{\mathrm{tg}}\alpha-1)\\ \mathop{\mathrm{tg}}\alpha=\dfrac{\sqrt2+1}{\sqrt2-1}=3+2\sqrt2=t](https://tex.z-dn.net/?f=%5Cmathop%7B%5Cmathrm%7Btg%7D%7D%5Cleft%28%5Calpha%2B%5Cdfrac%5Cpi4%5Cright%29%3D%5Cdfrac%7B%5Cmathop%7B%5Cmathrm%7Btg%7D%7D%5Calpha%2B%5Cmathop%7B%5Cmathrm%7Btg%7D%7D%5Cfrac%5Cpi4%7D%7B1-%5Cmathop%7B%5Cmathrm%7Btg%7D%7D%5Calpha%5Cmathop%7B%5Cmathrm%7Btg%7D%7D%5Cfrac%5Cpi4%7D%3D%5Cdfrac%7B1%2B%5Cmathop%7B%5Cmathrm%7Btg%7D%7D%5Calpha%7D%7B1-%5Cmathop%7B%5Cmathrm%7Btg%7D%7D%5Calpha%7D%3D-%5Csqrt2%5C%5C%0A1%2B%5Cmathop%7B%5Cmathrm%7Btg%7D%7D%5Calpha%3D%5Csqrt%7B2%7D%28%5Cmathop%7B%5Cmathrm%7Btg%7D%7D%5Calpha-1%29%5C%5C%0A%5Cmathop%7B%5Cmathrm%7Btg%7D%7D%5Calpha%3D%5Cdfrac%7B%5Csqrt2%2B1%7D%7B%5Csqrt2-1%7D%3D3%2B2%5Csqrt2%3Dt)
Формулы тангенса половинного угла:
![\cos2\alpha=\dfrac{1-t^2}{1+t^2}\qquad\sin2\alpha=\dfrac{2t}{1+t^2}](https://tex.z-dn.net/?f=%5Ccos2%5Calpha%3D%5Cdfrac%7B1-t%5E2%7D%7B1%2Bt%5E2%7D%5Cqquad%5Csin2%5Calpha%3D%5Cdfrac%7B2t%7D%7B1%2Bt%5E2%7D)
Тогда
![\sin4\alpha=2\sin2\alpha\cos2\alpha=\dfrac{4t(1-t^2)}{(1+t^2)^2}](https://tex.z-dn.net/?f=%5Csin4%5Calpha%3D2%5Csin2%5Calpha%5Ccos2%5Calpha%3D%5Cdfrac%7B4t%281-t%5E2%29%7D%7B%281%2Bt%5E2%29%5E2%7D)
![t^2=(3+2\sqrt2)^2=17+12\sqrt2\\ (1+t^2)^2=(18+12\sqrt2)^2=36(3+2\sqrt2)^2\\ \sin4\alpha=\dfrac{-4(3+2\sqrt2)(16+12\sqrt2)}{36(3+2\sqrt2)^2}=-\dfrac{4(4+3\sqrt2)}{9(3+2\sqrt2)}=\\=-\dfrac{4(4+3\sqrt2)(3-2\sqrt2)}{9}=-\dfrac{4\sqrt2}9](https://tex.z-dn.net/?f=t%5E2%3D%283%2B2%5Csqrt2%29%5E2%3D17%2B12%5Csqrt2%5C%5C%0A%281%2Bt%5E2%29%5E2%3D%2818%2B12%5Csqrt2%29%5E2%3D36%283%2B2%5Csqrt2%29%5E2%5C%5C%0A%5Csin4%5Calpha%3D%5Cdfrac%7B-4%283%2B2%5Csqrt2%29%2816%2B12%5Csqrt2%29%7D%7B36%283%2B2%5Csqrt2%29%5E2%7D%3D-%5Cdfrac%7B4%284%2B3%5Csqrt2%29%7D%7B9%283%2B2%5Csqrt2%29%7D%3D%5C%5C%3D-%5Cdfrac%7B4%284%2B3%5Csqrt2%29%283-2%5Csqrt2%29%7D%7B9%7D%3D-%5Cdfrac%7B4%5Csqrt2%7D9)
______________________________________________________
Другой способ. Найдём sin и cos, потом удвоим угол.
![\begin{cases}\cos^2\left(\alpha+\dfrac\pi4\right)=\dfrac1{1+\mathop{\mathrm{tg}}^2(\alpha+\frac\pi4)}=\dfrac13\\ \sin^2\left(\alpha+\dfrac\pi4\right)=1-\cos^2\left(\alpha+\dfrac\pi4\right)=\dfrac23\end{cases}\\ \begin{cases}-\sin2\alpha=\cos\left(2\alpha+\dfrac\pi2\right)=2\cos^2\left(\alpha+\dfrac\pi4\right)-1=-\dfrac13\\ \cos^22\alpha=\sin^2\left(2\alpha+\dfrac\pi2\right)=4\sin^2\left(\alpha+\dfrac\pi4\right)\cos^2\left(\alpha+\dfrac\pi4\right)=\dfrac89\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%5Ccos%5E2%5Cleft%28%5Calpha%2B%5Cdfrac%5Cpi4%5Cright%29%3D%5Cdfrac1%7B1%2B%5Cmathop%7B%5Cmathrm%7Btg%7D%7D%5E2%28%5Calpha%2B%5Cfrac%5Cpi4%29%7D%3D%5Cdfrac13%5C%5C%0A%5Csin%5E2%5Cleft%28%5Calpha%2B%5Cdfrac%5Cpi4%5Cright%29%3D1-%5Ccos%5E2%5Cleft%28%5Calpha%2B%5Cdfrac%5Cpi4%5Cright%29%3D%5Cdfrac23%5Cend%7Bcases%7D%5C%5C%0A%5Cbegin%7Bcases%7D-%5Csin2%5Calpha%3D%5Ccos%5Cleft%282%5Calpha%2B%5Cdfrac%5Cpi2%5Cright%29%3D2%5Ccos%5E2%5Cleft%28%5Calpha%2B%5Cdfrac%5Cpi4%5Cright%29-1%3D-%5Cdfrac13%5C%5C%0A%5Ccos%5E22%5Calpha%3D%5Csin%5E2%5Cleft%282%5Calpha%2B%5Cdfrac%5Cpi2%5Cright%29%3D4%5Csin%5E2%5Cleft%28%5Calpha%2B%5Cdfrac%5Cpi4%5Cright%29%5Ccos%5E2%5Cleft%28%5Calpha%2B%5Cdfrac%5Cpi4%5Cright%29%3D%5Cdfrac89%5Cend%7Bcases%7D)
Если tg(α + π/4) < 0, то πn + π/2 < α + π/4 < πn + π, πn + π/4 < α < πn + 3π4. Значит, 2πn + π/2 < 2α < 2πn + 3π/2, и cos 2α < 0.
![\begin{cases}\sin2\alpha=\dfrac13\\ \cos2\alpha=-\dfrac{2\sqrt2}3\end{cases}\\ \sin4\alpha=2\sin2\alpha\cos2\alpha=-\dfrac{4\sqrt2}9](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%5Csin2%5Calpha%3D%5Cdfrac13%5C%5C+%5Ccos2%5Calpha%3D-%5Cdfrac%7B2%5Csqrt2%7D3%5Cend%7Bcases%7D%5C%5C%0A%5Csin4%5Calpha%3D2%5Csin2%5Calpha%5Ccos2%5Calpha%3D-%5Cdfrac%7B4%5Csqrt2%7D9)
Уравнение линейной функции в общем виде выглядит так:
y=ax+b
Коэффициент b равен в первом случае 4,8
Нам нужно найти коэффициенты a
Если график проходит через точку (1,3), значит вместо x можно подставить 1, вместо y подставляем 3
3=а*1+4,8
а= -1,8
Готовое уравнение: y= -1,8x+4,8
2), 3), 4) так же, только коэффициент b меняется