Giải thích các bước giải:
Đặt $A=(x-1)(x+5)(x^2+4x+8)$
$→$ $A=(x^2+4x-5)(x^2+4x+8)$
Giả sử $t=x^2+4x-5$
$→A=t(t+13)=t^2+13t$
$→A= t^2 + 13t $ $+$ $\dfrac{169}{4}$ $-$ $\dfrac{169}{4}$
$→A=$ $(t$ $+$ $\dfrac{13}{2})^2$ $-$ $\dfrac{169}{4}$
Có $(t+$ $\dfrac{13}{2})^2≥0→A≤$ $\dfrac{169}{4}$
Dấu $"=$ xảy ra $⇔$ $(t$ $+$ $\dfrac{13}{2})^2=0→ t=$ $\dfrac{-13}{2}$
$→ x^2+4x-5=$ $\dfrac{-13}{2}$
$→ x^2+4x+4=$ $\dfrac{5}{2}$
$→ (x+2)^2=$ $\dfrac{5}{2}$
$→ x+2= ±\sqrt{\dfrac{5}{2}}$
$→$ \(\left[ \begin{array}{l}x+2=\sqrt{\dfrac{5}{2}}\\x+2=-\sqrt{\dfrac{5}{2}}\end{array} \right.\) $→$ \(\left[ \begin{array}{l}x=\dfrac{-4+\sqrt{10}}{2}\\x=\dfrac{-4-\sqrt{10}}{2}\end{array} \right.\)
Vậy $Max$ $A=\dfrac{169}{4}⇔$ $ \left[ \begin{array}{l}x=\dfrac{-4+\sqrt{10}}{2}\\x=\dfrac{-4-\sqrt{10}}{2}\end{array} \right.$
