Объяснение:

х=1 ; у= -4

(2у²+х)/(х-2у)=(2•(-4)²+1)/(1-2•(-4))=

=(2•16+1)/(1+8)=(32+1)/9=33/9=11/3=3 2/3

[tex]\displaystyle\bf\\ \ \ x=1 \ \ ; \ \ y=-4\\\\\\\frac{2y^{2} +x}{x-2y}=\frac{2\cdot(-4)^{2}+1 }{1-2\cdot(-4)} =\frac{2\cdot 16+1}{1+8} =\frac{32+1}{9} =\frac{33}{9} =3\frac{2}{3}[/tex]