Question

dau 25 de punce. efectuati reducind termenii asemenea
(x+3)²+(x+$\sqrt{3}$)²=
(3a-b)²-3(a-3b)+2(a+2b)²=
(2x+3)²-(x-5)²-4x(x-1)+x²+1=

Answer 1

[tex]a)(3a-b)^2-3(a-3b)+2(a+2b)^2=\\ 9a^2-6ab+b^2-3a+9b+2(a^2+2ab+b^2)=\\ 9a^2-6ab+b^2-3a+9b+2a^2+4ab+2b^2=11a^2-2ab+3b^2-3a+\\ 9b\\ b)(2x+3)^2-(x-5)^2-4x(x-1)+x^2+1=\\ 4x^2+12x+9-x^2+10x-25-4x^2+4x+x^2+1=26x-15[/tex]