Đáp án:
d. \( - 4{x^3} - \dfrac{{15}}{2}{x^2} - 10x - 3\)
Giải thích các bước giải:
\(\begin{array}{l}
a.F\left( x \right) = - {x^3} - \dfrac{1}{4}{x^2} + \dfrac{1}{2}x - x + 3{x^3} + 1\\
= 2{x^3} - \dfrac{1}{4}{x^2} - \dfrac{1}{2}x + 1\\
G\left( x \right) = 4{x^2} - 6 + 4x + {x^3} + 3x - 3{x^2} + 6\\
= {x^3} + {x^2} + 7x\\
H\left( x \right) = {x^3} - 3 + 6{x^2} + 8 - 6x + 8x\\
= {x^3} + 6{x^2} + 2x + 5\\
b.F\left( x \right) + G\left( x \right) = 2{x^3} - \dfrac{1}{4}{x^2} - \dfrac{1}{2}x + 1 + {x^3} + {x^2} + 7x\\
= 3{x^3} + \dfrac{3}{4}{x^2} + \dfrac{{13}}{2}x + 1\\
F\left( x \right) - G\left( x \right) = 2{x^3} - \dfrac{1}{4}{x^2} - \dfrac{1}{2}x + 1 - {x^3} - {x^2} - 7x\\
= {x^3} - \dfrac{5}{4}{x^2} - \dfrac{{15}}{2}x + 1\\
c.F\left( x \right) + G\left( x \right) + H\left( x \right)\\
= 3{x^3} + \dfrac{3}{4}{x^2} + \dfrac{{13}}{2}x + 1 + {x^3} + 6{x^2} + 2x + 5\\
= 4{x^3} + \dfrac{{27}}{4}{x^2} + \dfrac{{17}}{2}x + 6\\
d.2F\left( x \right) - G\left( x \right) - H\left( x \right)\\
= - 2{x^3} - \dfrac{1}{2}{x^2} - x + 2 - {x^3} - {x^2} - 7x - {x^3} - 6{x^2} - 2x - 5\\
= - 4{x^3} - \dfrac{{15}}{2}{x^2} - 10x - 3
\end{array}\)
