Đáp án:
$\begin{array}{l}
B4)\\
\left\{ \begin{array}{l}
f\left( x \right) + g\left( x \right) = 6{x^4} - 3{x^2} - 5\\
f\left( x \right) - g\left( x \right) = 4{x^4} - 6{x^3} + 7{x^2} + 8x - 9
\end{array} \right.\\
f\left( x \right) + g\left( x \right) + f\left( x \right) - g\left( x \right)\\
= 6{x^4} - 3{x^2} - 5 + 4{x^4} - 6{x^3} + 7{x^2} + 8x - 9\\
\Leftrightarrow 2f\left( x \right) = 10{x^4} - 6{x^3} + 4{x^2} + 8x - 14\\
\Leftrightarrow f\left( x \right) = 5{x^4} - 3{x^3} + 2{x^2} + 4x - 7\\
\Leftrightarrow g\left( x \right) = 6{x^4} - 3{x^2} - 5 - \left( {5{x^4} - 3{x^3} + 2{x^2} + 4x - 7} \right)\\
\Leftrightarrow g\left( x \right) = {x^4} + 3{x^3} - 5{x^2} - 4x + 2\\
Vay\,\left\{ \begin{array}{l}
f\left( x \right) = 5{x^4} - 3{x^3} + 2{x^2} + 4x - 7\\
g\left( x \right) = {x^4} + 3{x^3} - 5{x^2} - 4x + 2
\end{array} \right.\\
B5)\\
a)f\left( { - 1} \right) = 5.{\left( { - 1} \right)^3} - 7.{\left( { - 1} \right)^2} - 1 - 7\\
= - 5 - 7 - 1 - 7 = - 20\\
g\left( { - \dfrac{1}{2}} \right) = 7.{\left( { - \dfrac{1}{2}} \right)^3} - 7.{\left( { - \dfrac{1}{2}} \right)^2} + 2.\left( { - \dfrac{1}{2}} \right) + 5\\
= - \dfrac{7}{8} - \dfrac{7}{4} - 1 + 5\\
= - \dfrac{7}{8} - \dfrac{7}{4} + 4\\
= \dfrac{{11}}{8}\\
h\left( 0 \right) = {2.0^3} + 4.0 + 1 = 1\\
b)\\
k\left( x \right) = f\left( x \right) - g\left( x \right) + h\left( x \right)\\
= 5{x^3} - 7{x^2} + x - 7\\
- \left( {7{x^3} - 7{x^2} + 2x + 5} \right)\\
+ 2{x^3} + 4x + 1\\
= 3x - 11\\
c)k\left( x \right) = 3x - 11 = 0\\
\Leftrightarrow x = \dfrac{{11}}{3}
\end{array}$
Vậy k(x) bậc 1 và có nghiệm $x = \dfrac{{11}}{3}$
