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)`
