Đáp án:
b) \(H\left( x \right) = \dfrac{{13}}{2}x + \dfrac{{26}}{5}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)F\left( x \right) = 8{x^4} + \left( {4 - 2} \right){x^3} - \dfrac{1}{2}x + 5\\
= 8{x^4} + 2{x^3} - \dfrac{1}{2}x + 5\\
G\left( x \right) = - 8{x^4} + \left( {8 - 10} \right){x^3} + 7x + \dfrac{1}{5}\\
= - 8{x^4} - 2{x^3} + 7x + \dfrac{1}{5}\\
b)H\left( x \right) = F\left( x \right) + G\left( x \right)\\
= 8{x^4} + 2{x^3} - \dfrac{1}{2}x + 5 - 8{x^4} - 2{x^3} + 7x + \dfrac{1}{5}\\
= \dfrac{{13}}{2}x + \dfrac{{26}}{5}\\
P\left( x \right) = F\left( x \right) - G\left( x \right)\\
= 8{x^4} + 2{x^3} - \dfrac{1}{2}x + 5 + 8{x^4} + 2{x^3} - 7x - \dfrac{1}{5}\\
= 16{x^4} + 4{x^3} - \dfrac{{15}}{2}x + \dfrac{{24}}{5}\\
c)H\left( x \right) = 0\\
\to \dfrac{{13}}{2}x + \dfrac{{26}}{5} = 0\\
\to x = - \dfrac{4}{5}\\
R\left( x \right) = {x^2} + 9x + 8 = 0\\
\to {x^2} + x + 8x + 8 = 0\\
\to x\left( {x + 1} \right) + 8\left( {x + 1} \right) = 0\\
\to \left( {x + 1} \right)\left( {x + 8} \right) = 0\\
\to \left[ \begin{array}{l}
x = - 1\\
x = - 8
\end{array} \right.
\end{array}\)
