Bài 1:
$a.A=x^{2}+x-2$
$=(x^{2}+x+$ $\dfrac{1}{4})-$ $\dfrac{9}{4}$
$=(x+\dfrac{1}{2})^{2}-\dfrac{9}{4}$
Ta có: $(x+\dfrac{1}{2})^{2}$ $\geq0$
-> $(x+\dfrac{1}{2})^{2}-\dfrac{9}{4}$ $\geq\dfrac{-9}{4}$
Dấu bằng xảy ra khi: $x^{}+\dfrac{1}{2}=0$
⇔ $x^{}=\dfrac{-1}{2}$
$b.B=x^{2}+y^2+x-6y+5$
$=(x^{2}+x+\dfrac{1}{4})+(y^2-6x+9)-\dfrac{17}{4}$
$=(x+\dfrac{1}{2})^{2}+(y-3)^2-\dfrac{17}{4}$
Ta có: $(x+\dfrac{1}{2})^{2}+(y-3)^2$ $\geq0$
-> $(x+\dfrac{1}{2})^{2}+(y-3)^2-\dfrac{17}{4}$ $\geq\dfrac{-17}{4}$
Dấu bằng xảy ra khi: $\begin{cases} x+\dfrac{1}{2}=0\\\\y-3=0 \end{cases}$
⇔ $\begin{cases} x=\dfrac{-1}{2}\\\\y=3 \end{cases}$
Bài 2:
$e.E=-x^{2}-4y^2+2x-4y+3$
$=-(x^{2}+4y^2-2x+4y-3)$
$=-[(x^{2}-2x+1)+(4y^2+4y+1)-5]$
$=-[(x-1)^{2}+(2y+1)^2]+5$
Ta có: $(x-1)^{2}+(2y+1)^2$ $\geq0$
-> $-[(x-1)^{2}+(2y+1)^2]$ $\leq0$
-> $-[(x-1)^{2}+(2y+1)^2]+5$ $\leq5$
Dấu bằng xảy ra khi : $\begin{cases} x-1=0\\\\2y+1=0 \end{cases}$
⇔ $\begin{cases} x=1\\\\y=\dfrac{-1}{2} \end{cases}$
Chúc bạn học tốt !!!!!
