Đáp án:
c) \(\left[ \begin{array}{l}
x = 0\\
x = \sqrt 3 \\
x = - \sqrt 3
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)f\left( x \right) = 3{x^4} - {x^3} - x - 3{x^5}\\
= - 3{x^5} + 3{x^4} - {x^3} - x\\
g\left( x \right) = {x^4} + 2{x^2} + {x^5} - 2{x^3} - 7x + 3\\
= {x^5} + {x^4} - 2{x^3} + 2{x^2} - 7x + 3\\
h\left( x \right) = - 2{x^3} + 2{x^4} - {x^5} - \dfrac{5}{2}x + \dfrac{3}{2} + {x^2}\\
= - {x^5} + 2{x^4} - 2{x^3} + {x^2} - \dfrac{5}{2}x + \dfrac{3}{2}\\
b)A\left( x \right) = f\left( x \right) + g\left( x \right) - 2h\left( x \right)\\
= - 3{x^5} + 3{x^4} - {x^3} - x + {x^5} + {x^4} - 2{x^3} + 2{x^2} - 7x + 3 - 2\left( { - {x^5} + 2{x^4} - 2{x^3} + {x^2} - \dfrac{5}{2}x + \dfrac{3}{2}} \right)\\
= - 2{x^5} + 4{x^4} - 3{x^3} + 2{x^2} - 8x + 3 + 2{x^5} - 4{x^4} + 4{x^3} - 2{x^2} + 5x - 3\\
= {x^3} - 3x\\
c)A\left( x \right) = 0\\
\to {x^3} - 3x = 0\\
\to x\left( {{x^2} - 3} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = \sqrt 3 \\
x = - \sqrt 3
\end{array} \right.
\end{array}\)
