Đáp án:
$x=\pm \dfrac{2}{3}$
Giải thích các bước giải:
$F(x)-R(x)=3P(x)-6x^4+\dfrac{5}{2}x^3-\dfrac{2}{3}x^2+6x-4\\
\Rightarrow F(x)=3P(x)+R(x)-6x^4+\dfrac{5}{2}x^3-\dfrac{2}{3}x^2+6x-4\\
=3\left (\dfrac{3}{2}x^4-\dfrac{2}{3}x+3x^2-\dfrac{5}{6}x^3 \right )+\dfrac{3}{2}x^4+\dfrac{2}{3}x^2-4x-6x^4+\dfrac{5}{2}x^3-\dfrac{2}{3}x^2+6x-4\\
=\dfrac{9}{2}x^4-\dfrac{2.3}{3}x+9x^2-\dfrac{5.3}{6}x^3 +\dfrac{3}{2}x^4+\dfrac{2}{3}x^2-4x-6x^4+\dfrac{5}{2}x^3-\dfrac{2}{3}x^2+6x-4\\
=\left (\dfrac{9}{2}x^4 +\dfrac{3}{2}x^4 -6x^4 \right )+\left (-\dfrac{2.3}{3}x -4x+6x \right )+\left (9x^2 +\dfrac{2}{3}x^2 -\dfrac{2}{3}x^2\right )+\left (-\dfrac{5.3}{6}x^3 +\dfrac{5}{2}x^3 \right )-4\\
=\left (\dfrac{9+3}{2}x^4 -6x^4 \right )+\left (-2x +2x \right )+9x^2 +\left (-\dfrac{5}{2}x^3 +\dfrac{5}{2}x^3 \right )-4\\
=\left (6x^4 -6x^4 \right )+9x^2 -4\\
=9x^2 -4\\
\Rightarrow F(x)=0\\
\Leftrightarrow 9x^2-4=0\\
\Leftrightarrow 9x^2=4\\
\Leftrightarrow x^2=\dfrac{4}{9}\\
\Leftrightarrow x=\pm \sqrt{\dfrac{4}{9}}=\pm \dfrac{2}{3}$
