Question

Rezultatul calculului (2x+1)^2−(3x−1)(2x+1)+(x−2)^2 este:

Answer 1

(2x+1)²-(3x-1)(2x+1)+(x-2)²=
=[(2x)²+2*2x*1+1²]-(3x*2x+3x*1-2x*1-1*1)-(x²-2*2*x+2²)=
=4x²+4x+1-(6x²+3x-2x-1)-x²+4x-4=
=4x²+4x+1-6x²-x+1-x²+4x-4=
=-3x²+7x-2

Answer 2

$(2x+1)^2-(3x-1)(2x+1)+(x-2)^2 = \\ =(2x)^2+2 \cdot 2x \cdot 1+1^2-(6x^2+3x-2x-1)+x^2-2 \cdot x \cdot 2+2^2= \\ =4x^2+4x+1-6x^2-3x+2x+1+x^2-4x+4=-x^2-x+6$