Cho 2 số thực dương x,y thỏa mãn
$\frac{y}{2x+3}$ =$\frac{\sqrt[]{2x+3}+1 }{\sqrt[]{y}+1 }$
tìm GTNN của Q=xy-3y-2x-3
Cho 2 số thực dương x,y thỏa mãn
$\frac{y}{2x+3}$ =$\frac{\sqrt[]{2x+3}+1 }{\sqrt[]{y}+1 }$
tìm GTNN của Q=xy-3y-2x-3
Đáp án: `Q_{min}=-\frac{121}{8}⇔(x;y)=(\frac{5}{4};\frac{11}{2})`
Giải thích các bước giải:
`ĐKXĐ: x>\frac{-3}{2}; y≥0`
Ta có: `\frac{y}{2x+3}=\frac{\sqrt{2x+3}+1}{\sqrt{y}+1}`
$⇔y(\sqrt{y}+1)=(2x+3)(\sqrt{2x+3}+1)$
$⇔y\sqrt{y}+y=(2x+3)\sqrt{2x+3}+(2x+3)$
$⇔y\sqrt{y}-(2x+3)\sqrt{2x+3}+y-(2x+3)=0$
$⇔(\sqrt{y})^3-(\sqrt{2x+3})^3+(\sqrt{y})^2-(\sqrt{2x+3})^2=0$
$⇔(\sqrt{y}-\sqrt{2x+3})[(\sqrt{y})^2+\sqrt{y}\sqrt{2x+3}+(\sqrt{2x+3})^2]+(\sqrt{y}-\sqrt{2x+3})(\sqrt{y}+\sqrt{2x+3})=0$
$⇔(\sqrt{y}-\sqrt{2x+3})[(\sqrt{y})^2+\sqrt{y}\sqrt{2x+3}+(\sqrt{2x+3})^2+\sqrt{y}+\sqrt{2x+3}]=0$
$⇔\sqrt{y}-\sqrt{2x+3}=0$ (do $(\sqrt{y})^2+\sqrt{y}\sqrt{2x+3}+(\sqrt{2x+3})^2+\sqrt{y}+\sqrt{2x+3}>0$)
$⇔\sqrt{y}=\sqrt{2x+3}⇔y=2x+3$
Ta có: $Q=xy-3y-2x-3$
$=x(2x+3)-3(2x+3)-2x-3$
$=2x^2+3x-6x-9-2x-3$
$=2x^2-5x-12$
`=2(x^2-2.x.\frac{5}{4}+\frac{25}{16})-\frac{121}{8}`
`=2(x-\frac{5}{4})^2-\frac{121}{8}`
Do `(x-\frac{5}{4})^2≥0`
`⇒2(x-\frac{5}{4})^2≥0`
`⇒Q=2(x-\frac{5}{4})^2-\frac{121}{8}≥-\frac{121}{8}`
Dấu bằng xảy ra `⇔(x-\frac{5}{4})^2=0`
`⇔x-\frac{5}{4}=0⇔x=\frac{5}{4}`
`⇒y=2.\frac{5}{4}+3=\frac{11}{2}`