Odpowiedź:
[tex]a)\ \ (x-2)^2+(x+1)\cdot3=x^2-4x+4+3x+3=x^2-x+7\\\\\\b)\ \ (4a+2)^2-(a-1)^2=16a^2+16a+4-(a^2-2a+1)=16a^2+16a+4-a^2+2a-1=\\\\=15a^2+18a+3\\\\\\c)\ \ (2a-b)(2a+b)+2(a^2+3ab-b^2)=4a^2-b^2+2a^2+6ab-2b^2=6a^2-3b^2+6ab\\\\\\d)\ \ 2(a+1)^2-3(a-2)^2=2(a^2+2a+1)-3(a^2-4a+4)=\\\\=2a^2+4a+2-3a^2+12a-12=-a^2+16a-10\\\\\\e)\ \ (x+y)(2x-y)+(2x-y)^2=2x^2-xy+2xy-y^2+(4x^2-4xy+y^2)=\\\\=2x^2+xy-y^2+4x^2-4xy+y^2=6x^2-3xy[/tex]
[tex]f)\ \ (3a-2b)^2-(a+2b)^2-(a-b)(a+b)=\\\\=9a^2-12ab+4b^2-(a^2+4ab+4b^2)-(a^2-b^2)=\\\\=9a^{2}-12ab+4b^{2}-a^{2}-4ab-4b^{2}-a^{2}+b^{2}=7a^{2}-16ab+b^{2}[/tex]
[tex]Wykorzystane\ \ wzory\\\\(a-b)^2=a^{2}-2ab+b^{2}\\\\(a+b)^2=a^{2}+2ab+b^{2}\\\\(a-b)(a+b)=a^{2}-b^{2}[/tex]
