Объяснение:

[tex]1)\ 3*log_52+2-x=log_5(3^x-5^{2-x})\\log_52^3+log_55^2-log_55^x=log_5(3^x-5^{2-x})\\log_5\frac{8*25}{5^x}=log_5(3x-5^{2-x})\\ \frac{33}{5^x}=3^x-5^{2-x}\\ \frac{200}{5^x} =3^x-\frac{5^2}{5^x}\\3^x= \frac{200}{5^x}+\frac{25}{5^x} \\3^x=\frac{225}{5^x}\\ 3^x*5^3=225\\(3*5)^x=225\\15^x=15^2\\x=2.[/tex]

[tex]2)\ \left \{ {{y*x^{log_yx}=x^{2,5}} \atop {log_3y*log_y(y-2x)=1}} \right. \\[/tex]

ОДЗ: х≥0    у≥0    у≠1 .

[tex]y*x^{log_yx}=x^{2.5}.\\log_x(y*x^{log_yx})}=log_xx^{2,5}\\log_xy+log_xx^{log_yx}=2,5*log_xx\\log_xy+log_yx*log_xx=2,5\\log_xy+log_yx=2,5\\log_xy+\frac{1}{log_xy}=2,5\\ log^2_xy+1=2,5*log_xy\\log^2_xy-2,5log_xy+1=0.[/tex]

Пусть logₓy=t.        ⇒

[tex]t^2-2,5t+1=0\\D=2,25\ \ \ \ \ \\sqrt{D}=1,5\\ t_1=0,5\ \ \ \ t_2=2.\\t_1=log_xy=0,5\\y_1=x^{0,5}=\sqrt{x}. \\y_2=x^2.[/tex]

[tex]log_3y*log_y(y-2x)=1\\log_y(y-2x)=\frac{1}{log_3y}=log_y3\\ y-2x=3\\y=2x-3.\\[/tex]

[tex]1)\ \left \{ {{y=\sqrt{x} } \atop {y=2x+3}} \right. \ \ \ \ \Rightarrow\\\sqrt{x} =2x+3\\(\sqrt{x})^2=(2x+3)^2\\x=4x^2+12x+9\\4x^2+11x+9=0\\D=-23\ \ \ \ \Rightarrow\\x\in \varnothing.\\2)\ \left \{ {{y=x^2} \atop {y=2x+3}} \right. \ \ \ \ \ \Rightarrow\ \ \ \\x^2=2x+3\\x^2-2x-3=0\\D=16\ \ \ \ \sqrt{D}=4\\ x_1=-1\notin\\ x_2=3\in\ \ \ \ \ y=2*3+3=6+3=9.[/tex]

Ответ: (3;9).