![\left \{ {{4^{2x-y}=64} \atop {5^{x-y}=5}} \right. \to \left \{ {{4^{2x-y}=4^3} \atop {5^{x-y}=5^1}} \right. \to \left \{ {{2x-y=3} \atop {x-y=1}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7B4%5E%7B2x-y%7D%3D64%7D+%5Catop+%7B5%5E%7Bx-y%7D%3D5%7D%7D+%5Cright.+%5Cto++%5Cleft+%5C%7B+%7B%7B4%5E%7B2x-y%7D%3D4%5E3%7D+%5Catop+%7B5%5E%7Bx-y%7D%3D5%5E1%7D%7D+%5Cright.+%5Cto++%5Cleft+%5C%7B+%7B%7B2x-y%3D3%7D+%5Catop+%7Bx-y%3D1%7D%7D+%5Cright.+)
Из уравнения 2 выразим переменную х
x = 1+y, и подставим вместо х в 1-е уравннеие
![2(y+1)-y=3\\ 2y+2-y=3\\ y=1 \\ x=y+1=2](https://tex.z-dn.net/?f=2%28y%2B1%29-y%3D3%5C%5C+2y%2B2-y%3D3%5C%5C+y%3D1+%5C%5C+x%3Dy%2B1%3D2)
Окончательный ответ:
![(2;1)](https://tex.z-dn.net/?f=%282%3B1%29)
25-10m<0
-10m<-25
m>10/25
m>2/5
50^(<em>k + 3</em>) = 25^(<em>k + 3</em>)<em /> * 2^(<em>k + 3) = </em>5^2(<em>k + 3</em>)<em /> * 2^(<em>k + 3)
</em><em>
</em>5^2(<em>k + 3</em>)<em /> : 5^(<em>2k + 5</em>) * 2^(<em>k + 3) : 2</em>^(<em>k - 2</em>)<em> = 5 * 2 ^5 = 5 * 32 = 160</em><em>
</em><em>
</em>
(3+√5)(5-√5)(√5+1)²=(3+√5)·√5(√5-1)·(√5+1)·(√5+1)=√5(3+√5)·4·(√5+1)=
4√5·(3√5+3+5+√5)=4√5(4√5+8)=16√5(√5+2)=80+32√5.