Этих 4-ри угла являют собой две пары одинаковых углов, т.е.:
![2\alpha+2\beta=360^0\\\\ \alpha+\beta=180^0](https://tex.z-dn.net/?f=2%5Calpha%2B2%5Cbeta%3D360%5E0%5C%5C%5C%5C%0A%5Calpha%2B%5Cbeta%3D180%5E0)
известно, что
![2\alpha+\beta=325^0\\\\ \beta=325^0-2\alpha](https://tex.z-dn.net/?f=2%5Calpha%2B%5Cbeta%3D325%5E0%5C%5C%5C%5C%0A%5Cbeta%3D325%5E0-2%5Calpha)
тогда
![\alpha+\beta=180^0\\\\ \alpha+325^0-2\alpha=180^0\\\\ -\alpha=180^0-325^0\\\\ -\alpha=-145^0\\\\ \alpha=145^0\\\\ \beta=180^0-145^0=35^0](https://tex.z-dn.net/?f=%5Calpha%2B%5Cbeta%3D180%5E0%5C%5C%5C%5C%0A%5Calpha%2B325%5E0-2%5Calpha%3D180%5E0%5C%5C%5C%5C%0A-%5Calpha%3D180%5E0-325%5E0%5C%5C%5C%5C%0A-%5Calpha%3D-145%5E0%5C%5C%5C%5C%0A%5Calpha%3D145%5E0%5C%5C%5C%5C%0A%5Cbeta%3D180%5E0-145%5E0%3D35%5E0)
Ответ:
![145^0;\ \ 35^0](https://tex.z-dn.net/?f=145%5E0%3B%5C+%5C+35%5E0)
2^10(2*5)=1024
3^10=59049
6^7=279936
1024*59049/279936=216
1-ый способ: 56am+44mu−32au−77m²=(56am-77m<span>²)+(44mu-32au)=7m(8a-11m)+4u(11m-8a)=7m(8a-11m)-4u(8a-11)=(7m-4u)(8a-11m)
2-ой способ: </span>56am+44mu−32au−77m²=(56am-32au)+(44mu-77m²)=8a(7m-4u)+11m(4u-7m)=8a(7m-4u)-11m(7m-4u)=(8a-11m)(7m-4u)