Với x,y không âm tìm GTNN của P= x - 2cănxy + 3y - 2cănx + 2009.5
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\(2P=2x-4\sqrt{xy}+6y-4\sqrt{x}+4019\)
\(=\left(\left(x-4\sqrt{xy}+y\right)-\frac{2}{2}.\left(\sqrt{x}-2\sqrt{y}\right)+\frac{1}{4}\right)+\left(x-\frac{2.3.\sqrt{x}}{2}+\frac{9}{4}\right)+2\left(y-\frac{2\sqrt{y}}{2}+\frac{1}{4}\right)+4016\)
\(=\left(\left(\sqrt{x}-2\sqrt{y}\right)^2-\frac{2}{2}.\left(\sqrt{x}-2\sqrt{y}\right)+\frac{1}{4}\right)+\left(x-\frac{2.3.\sqrt{x}}{2}+\frac{9}{4}\right)+2\left(y-\frac{2\sqrt{y}}{2}+\frac{1}{4}\right)+4016\)
\(=\left(\sqrt{x}-2\sqrt{y}-\frac{1}{2}\right)^2+\left(\sqrt{x}-\frac{3}{2}\right)^2+2\left(\sqrt{y}-\frac{1}{2}\right)^2+4016\ge2016\)
\(\Rightarrow P\ge2008\)khi \(\hept{\begin{cases}x=\frac{9}{4}\\y=\frac{1}{4}\end{cases}}\)
tung hỏa mù hả sao tăng Hệ số lên làm gì?
căn x=a, căn y=b
P=(a^2+b^2-2ab-2a+2b+1)+(2b^2-2b+1/2)+2009+1/2-(1+1/2)
P=(a-b-1)^2+2(b-1/2)^2+2008>=2008
đăng thức b=1/2=>y=1/4; và a-1/2-1=0=>a=3/2=>x=9/4
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\(P=x+y+\dfrac{10}{x+y}=2\sqrt{10}\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}x+y=\sqrt{10}\\x^2+y=8\end{matrix}\right.\)
\(\Rightarrow\left(x;y\right)=\left(\dfrac{1+\sqrt{33-4\sqrt{10}}}{2};\dfrac{2\sqrt{10}-1-\sqrt{33-4\sqrt{10}}}{2}\right)\)
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Đặt \(a=\sqrt{x},b=\sqrt{y}\) thì \(a,b\ge0\)
\(P=a^2-2ab+3b^2-2a+2004,5=\left(\frac{a^2}{3}-2ab+3b^2\right)+\left(\frac{2}{3}a^2-2a+\frac{3}{2}\right)+2003\)
\(=\left(\frac{a}{\sqrt{3}}-\sqrt{3}b\right)^2+\frac{2}{3}\left(a-\frac{3}{2}\right)^2+2003\ge2003\)
Dấu "=" xảy ra khi a = 3/2 , b = 1/2
Vậy Min P = 2003 khi x = 9/4 , y = 1/4
Đặt \(a=\sqrt{x},b=\sqrt{y}\) thì \(a,b\ge0\)
\(P=a^2-2ab+3b^2-2a+2004,5=\left(\frac{a^2}{3}-2ab+3b^2\right)+\left(\frac{2}{3}a^2-2a+\frac{3}{2}\right)+2003\)
\(=\left(\frac{a}{\sqrt{3}}-\sqrt{3}b\right)^2+\frac{2}{3}\left(a-\frac{3}{2}\right)^2+2003\ge2003\)
Dấu "=" xảy ra khi a = 3/2 , b = 1/2
Vậy Min P = 2003 khi x = 9/4 , y = 1/4
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Ta có: \(\left(\left|x-3\right|+2\right)^2\ge0\forall x\) không âm
\(\left|y+3\right|\ge3\forall y\) không âm
Cộng theo vế 2 BĐT trên ta có:
\(A=\left(\left|x-3\right|+2\right)^2+\left|y+3\right|+2018\ge0+3+2018=2021\)
Vậy \(A_{min}=2021\Leftrightarrow\hept{\begin{cases}\left(\left|x-3\right|+2\right)^2=0\\\left|y+3\right|=3\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=0\end{cases}}}\)