`a)`

`P(x)=5x³-3x+7-x`

        `=5x³-(3x+x)+7`

        `=5x³-4x+7`

`Q(x)=-5x³+2x-3+2x-x²-2`

        `=-5x³-x²+(2x+2x)-(3+2)`

        `=-5x³-x²+4x-5`

`M(x)=P(x)+Q(x)`

        `=5x³-4x+7-5x³-x²+4x-5`

        `=(5x³-5x³)-x²+(-4x+4x)+(7-5)`

        `=-x²+2`

`N(x)=P(x)-Q(x)`

        `=5x³-4x+7-(-5x³-x²+4x-5)`

        `=5x³-4x+8+5x³+x²-4x+5`

        `=(5x³+5x³)+x²-(4x+4x)+(7+5)`

        `=10x³+x²-8x+12`

`b)`

Đặt `M(x)=0⇒-x²+2=0`

    `-x²+2=0`

`→-x²=-2`

`→x²=2`

`→x=±`$\sqrt[]{2}$ 

Vậy `2` nghiệm của đa thức `M(x)` là `x=±`$\sqrt[]{2}$ 

Đáp án:

$\\$

`a,`

$\bullet$ `P (x) = 5x^3 - 3x + 7 -x`

`-> P(x) = 5x^3 + (-3x-x) + 7`

`-> P (x)=5x^3 - 4x + 7`

$\bullet$ `Q(x) = -5x^3 + 2x - 3 + 2x -x^2-2`

`-> Q (x) = -5x^3 + (2x+2x) -x^2 + (-3-2)`

`-> Q (x) = -5x^3 + 4x - x^2 - 5`

`-> Q (x) = -5x^3 - x^2 + 4x-5`

$\bullet$ `M (x)=P(x) + Q (x)`

`-> M (x) = (5x^3 - 4x + 7) + (-5x^3 - x^2 + 4x-5)`

`-> M (x) = 5x^3 - 4x + 7 -5x^3 - x^2 + 4x-5`

`-> M (x) = (5x^3 -5x^3) + (-4x+4x) - x^2 + (7-5)`

`->M (x) = -x^2 + 2`

$\bullet$ `N (x) = P (x)-Q (x)`

`-> N (x)  = (5x^3 - 4x + 7) - (-5x^3 - x^2 + 4x-5)`

`-> N (x) = 5x^3 - 4x + 7 + 5x^3 + x^2 - 4x +5`

`-> N (x) = (5x^3 + 5x^3) + (-4x-4x) + (7+5) + x^2`

`->  N (x) = 10x^3 - 8x + 12 + x^2`

$\\$

`b,`

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

Cho `M(x)=0`

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

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

`->-x^2=-2`

`->x^2=2`

`->` \(\left[ \begin{array}{l}x^2=(\sqrt{2})^2\\x^2=(-\sqrt{2})^2\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}x=\sqrt{2}\\x=-\sqrt{2}\end{array} \right.\) 

Vậy `x=\sqrt{2},x=-\sqrt{2}` là 2 nghiệm của `M (x)`