A = 2x.(x - 3) – x(x -7)- 5(x - 403)
= 2x2 - 6x - x2 + 7x - 5x + 2015
= x2 + x + 2015
x = 4 => A = 42 + 4 + 2015 = 2035
A = 2015 => x2 + x + 2015 = 2015
<=> x2 + x = 0
<=> x(x + 1) = 0
<=> x = 0 hoặc x = -1
Ta có \(A=2x^2-6x-x^2+7x-5x+2015\)
\(=x^2-4x+2015\)
A, Với \(x=4\) ta được \(A=2015\)
B, \(A=2015\Rightarrow x^2-4x=0\Rightarrow x\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)