Giải thích các bước giải:
$\text{Bài 1:}$
$P=3-x^2+2x$
$=-(x^2-2x-3)$
$=-(x^2-2x+1-4)$
$=-[(x-1)^2-4]$
$=-(x-1)^2+4$
$\text{Ta có:}$
$-(x-1)^2≤0$ $∀x∈R$
$⇒-(x-1)^2+4≤4$ $∀x∈R$
$\text{Dấu "=" xảy ra khi:}$
$-(x-1)^2+4=4$
$⇔-(x-1)^2=0$
$⇔x-1=0$
$⇔x=1$
$\text{Vậy $Max_{(P)}=4$ tại $x=1$}$
$\text{Bài 2:}$
$4x^2-(3x+1)^2=0$
$⇔(2x-3x-1)(2x+3x+1)=0$
$⇔(-x-1)(5x+1)=0$
$⇔-(x+1)(5x+1)=0$
$⇔(x+1)(5x+1)=0$
$⇔$ \(\left[ \begin{array}{l}x+1=0\\5x+1=0\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x=-1\\5x=-1\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x=-1\\x=-\dfrac{1}{5}\end{array} \right.\)
$\text{Vậy $x∈\{-1;-\dfrac{1}{5}\}$}$
Học tốt!!!
