\(a,B=3-2x+\sqrt{1+4x+4x^2}\\ =3-2x+\sqrt{\left(2x+1\right)^2}\\ =3-2x+2x+1\\ =4\)
\(b,\) Thay \(x=2015\) ta có:
\(B=4\)
a, B=3-2x+\(\sqrt{4x^2+4x+1}\) =3-2x+\(\sqrt{\left(2x+1\right)^2}\) =3-2x+2x+1=4 b; Ta co B=4=4+0x thay x=2015 =>B=4+0x=4+0.2015=4 vay x=2015=>B=4