$4$.
$A = 5 - (3x+2)^2$
Vì $(3x+2)^2 ≥ 0 ∀ x$ nên $A = 5 - (3x+2)^2 ≤ 5 ∀ x$
Dấu "$=$" xảy ra $⇔$ $(3x+2)^2 = 0 ⇔ x = \dfrac{-2}{3}$
Vậy $A_{max} = 5$ khi $x = \dfrac{-2}{3}$
$B = -6 - |2x-5|$
$⇔ B = - (6 + |2x-5|)$
Vì $|2x-5|≥ 0 ∀ x$ nên $B = - (6 + |2x-5|)≤ -6 ∀ x$
Dấu "$=$" xảy ra $⇔$ $|2x-5| = 0 ⇔ x = \dfrac{5}{2}$
Vậy $B_{max} = -6$ khi $x = \dfrac{5}{2}$
$5$.
Ta có: $P(x) = (x-2)^{49} - x+2$
$⇔ P(x) = (x-2)^{49} - (x-2)$
$⇔ P(x) = (x-2).[(x-2)^{48} - 1]$
$P(x) = 0 ⇒ (x-2).[(x-2)^{48} - 1]= 0$
$⇒$ \(\left[ \begin{array}{l}x-2=0\\(x-2)^{48}=1\end{array} \right.\)
$⇒$ \(\left[ \begin{array}{l}x=2\\x=3\\x=1\end{array} \right.\)
Vậy $P(x)$ có $3$ nghiệm $x=1$;$x=2$ và $x=3$.
