Odpowiedź:
[tex]4\cdot(2x+3y)^2-(4-y)^2=4(4x^2+12xy+9y^2)-(16-8y+y^2)=\\\\=16x^2+48xy+36y^2-16+8y-y^2=16x^2+48xy+35y^2+8y-16\\\\\\dla\ \ x=-1\ \ ,\ \ y=1\frac{1}{2}\\\\16x^2+48xy+35y^2+8y-16=16\cdot(-1)^2+48\cdot(-1)\cdot1\frac{1}{2}+35\cdot(1\frac{1}{2})^2+8\cdot1\frac{1}{2}-16=\\\\=16\cdot1-\not48^2^4\cdot\frac{3}{\not2_{1}}+35\cdot(\frac{3}{2})^2+\not8^4\cdot\frac{3}{\not2_{1} }-16=16-72+35\cdot\frac{9}{4}+12-16=[/tex]
[tex]-60+\frac{315}{4}=-60+78\frac{3}{4}=18\frac{3}{4}[/tex]