Đáp án:
$\begin{array}{l}
2)a)A\left( 1 \right) = {1^3} - {5.1^2} + 7.1 = 3\\
B\left( { - 1} \right) = - {\left( { - 1} \right)^4} - 11.\left( { - 1} \right) + {\left( { - 1} \right)^2} + 3\\
= - 1 + 11 + 1 + 3\\
= 14\\
C\left( { - \dfrac{1}{2}} \right) = - 4.\left( { - \dfrac{1}{2}} \right) + 5.{\left( { - \dfrac{1}{2}} \right)^3} + {\left( { - \dfrac{1}{2}} \right)^2} - 8\\
= 2 - \dfrac{5}{8} + \dfrac{1}{4} - 8\\
= \dfrac{{ - 3}}{8} - 6 = \dfrac{{ - 51}}{8}\\
b)A\left( x \right) + B\left( x \right)\\
= {x^3} - 5{x^2} + 7x - {x^4} - 11x + {x^2} + 3\\
= - {x^4} + {x^3} - 4{x^2} - 4x + 3\\
A\left( x \right) - B\left( x \right)\\
= {x^3} - 5{x^2} + 7x + {x^4} + 11x - {x^2} - 3\\
= {x^4} + {x^3} - 6{x^2} + 18x - 3\\
B\left( x \right) + C\left( x \right)\\
= - {x^4} - 11x + {x^2} + 3 - 4x + 5{x^3} + {x^2} - 8\\
= - {x^4} + 5{x^3} + 2{x^2} - 15x - 5\\
A\left( x \right) - C\left( x \right)\\
= {x^3} - 5{x^2} + 7x + 4x - 5{x^3} - {x^2} + 8\\
= - 4{x^3} - 6{x^2} + 11x + 8\\
c)D\left( x \right) - \left[ {B\left( x \right) - C\left( x \right)} \right] = A\left( x \right)\\
\Leftrightarrow D\left( x \right) = B\left( x \right) - C\left( x \right) + A\left( x \right)\\
= - {x^4} - 11x + {x^2} + 3 + 4x - 5{x^3} - {x^2} + 8\\
+ {x^3} - 5{x^2} + 7x\\
= - {x^4} - 4{x^3} - 5{x^2} + 11\\
3)a)x = - 1\\
\Leftrightarrow D\left( { - 1} \right) = 0\\
\Leftrightarrow m.{\left( { - 1} \right)^2} + 5.\left( { - 1} \right) - 3 = 0\\
\Leftrightarrow m - 8 = 0\\
\Leftrightarrow m = 8\\
Vậy\,m = 8\\
b)A\left( x \right) = 0\\
\Leftrightarrow {x^3} - 2x - \left( {x + {x^3} - 9} \right) = 0\\
\Leftrightarrow - x + 9 = 0\\
\Leftrightarrow x = 9\\
c)E\left( x \right) = 0\\
\Leftrightarrow {x^2} - 2x + 1 = 0\\
\Leftrightarrow {\left( {x - 1} \right)^2} = 0\\
\Leftrightarrow x = 1\\
d)B\left( 0 \right) = - {0^4} - 11.0 + {0^2}
\end{array}$
Vậy x=0 không là nghiệm của đa thức
