Odpowiedź:
[tex]Poziom\ \ D\\\\a)\ \ 2(x-8)-3(y-2x^2)=2x-16-3y+6x^2\\\\\\b)\ \ \frac{9a+3b}{3}-\frac{4a-8b}{2}+a=\frac{\not3(3a+b)}{\not3}-\frac{\not2(2a-4b)}{\not2}+a=3a+b-(2a-4b)+a=\\\\=3a+b-2a+4b+a=2a+5b\\\\\\c)\ \ -5(p-r+4)-(3r-5p)+16=-5p+5r-20-3r+5p+16=2r-4\\\\\\d)\ \ \frac{4t-8}{4}+(-6t+2s-1):2=\frac{\not4(t-2)}{\not4}-3t+s-0,5=t-2-3t+s-0,5=\\\\=-2t+s-2,5[/tex]
[tex]e)\ \ -6x-(3x-y)+4(-2x+y)=-6x-3x+y-8x+4y=-17x+5y\\\\\\f)\ \ \frac{12g+4}{2}-\frac{10g-1}{5}=\frac{\not2(6g+2)}{\not2}-\frac{\not5(2g-0,2)}{\not5}=6g+2-(2g-0,2)=6g+2-2g+0,2=\\\\=4g+2,2[/tex]
