Đáp án:

$\\$

$\bullet$ `M (x) = x - 4x^2 + 5x - 7`

`-> M(x) = (x + 5x)- 4x^2 - 7`

`-> M (x) = 6x - 4x^2 - 7`

`-> M (x) = -4x^2 + 6x - 7`

$\bullet$ `N (x) = -2x + 4x - 4x + 7`

`-> N (x) = (-2x + 4x - 4x) + 7`

`-> N (x) = -2x + 7`

$\\$

`a,`

$\bullet$ `A (x) = M (x) + N (x)`

`-> A (x) = (-4x^2 + 6x - 7) + (-2x + 7)`

`-> A (x) = -4x^2 + 6x - 7 - 2x + 7`

`->A (x)  = -4x^2 + (6x - 2x) + (-7 + 7)`

`-> A (x) = -4x^2 + 4x`

$\bullet$ `B (x) = M (x) - N (x)`

`-> B (x) = (-4x^2 + 6x - 7) - (-2x + 7)`

`-> B  (x) = -4x^2 + 6x - 7 + 2x - 7`

`-> B (x) = -4x^2  + (6x + 2x) + (-7 - 7)`

`-> B (x) = -4x^2 + 8x - 14`

$\\$

`b,`

$\bullet$ `A (x) = -4x^2 + 4x`

Cho `A (x) = 0`

`-> -4x^2 + 4x = 0`

`-> x (-4x + 4) = 0`

`->` \(\left[ \begin{array}{l}x=0\\-4x+4=0\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}x=0\\-4x=-4\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\) 

Vậy `x=0,x=1` là 2 nghiệm của `A (x)`