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\(\sqrt{x-1}-y\sqrt{y}=\sqrt{y-1}-x\sqrt{x}\)
\(\Leftrightarrow\left(\sqrt{x-1}-\sqrt{y-1}\right)+\left(x\sqrt{x}-y\sqrt{y}\right)=0\)
\(\Leftrightarrow\frac{x-y}{\sqrt{x-1}+\sqrt{y-1}}+\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)\left(\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x-1}+\sqrt{y-1}}+x+\sqrt{xy}+y\right)=0\)
\(\Leftrightarrow x=y\)
\(\Rightarrow S=2x^2-8x+5=2\left(x-2\right)^2-3\ge-3\)
Tại sao từ:\(\left(\sqrt{x-1}-\sqrt{y-1}\right)\) lại => đc: \(\frac{x-y}{\sqrt{x-1}+\sqrt{y-1}}\)??????????
1/ \(\frac{3}{2}x^2+y^2+z^2+yz=1\Leftrightarrow3x^2+2y^2+2z^2+2yz=2\)
\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2-2xy+y^2\right)+\left(x^2-2zx+z^2\right)=2\)
\(\Leftrightarrow\left(x+y+z\right)^2+\left(x-y\right)^2+\left(x-z\right)^2=2\)
\(\Rightarrow-\sqrt{2}\le x+y+z\le\sqrt{2}\)
Suy ra MIN A = \(-\sqrt{2}\)khi \(x=y=z=-\frac{\sqrt{2}}{3}\)
Từ điều kiện suy ra \(\sqrt{xy}+\sqrt{x}+\sqrt{y}\ge3\)
Áp dụng BĐT Cô-si, ta có :
\(3\le\sqrt{xy}+\sqrt{x}.1+\sqrt{y}.1\le\frac{x+y}{2}+\frac{x+1}{2}+\frac{y+1}{2}\)
\(\Rightarrow x+y\ge2\)
Ta có : \(\frac{x^2}{y}+y\ge2\sqrt{\frac{x^2}{y}.y}=2x\); \(\frac{y^2}{x}+x\ge2\sqrt{\frac{y^2}{x}.x}=2y\)
\(\Rightarrow\frac{x^2}{y}+\frac{y^2}{x}+x+y\ge2x+2y\)
\(\Rightarrow P=\frac{x^2}{y}+\frac{y^2}{x}\ge x+y\ge2\)
Vậy GTNN của P là 2 khi x = y = 1
\(\sqrt{x-1}-y\sqrt{y}=\sqrt{y-1}-x\sqrt{x}\)(ĐK:\(x;y\ge1\))
\(\Leftrightarrow\sqrt{x-1}+x\sqrt{x}=\sqrt{y-1}+y\sqrt{y}\)
Xét x<y\(\Rightarrow\sqrt{x-1}< \sqrt{y-1};x\sqrt{x}< y\sqrt{y}\)
\(\Rightarrow VT< VP\)
TT xét x>y=>VT>VP
\(\Rightarrow x=y\)
\(\Rightarrow S=x^2+3x^2-2x^2-8x+5\)
\(S=2x^2-8x+5=2\left(x-2\right)^2-3\ge-3\)
"="<=>x=y=2(tm)
Áp dụng BĐT Cô-si ta có:
\(2x^2+3xy+4y^2\ge3\sqrt[3]{2x^2\cdot3xy\cdot4y^2}=3\sqrt[3]{24x^3y^3}\Rightarrow\sqrt{2x^2+3xy+4y^2}\ge\sqrt{xy\cdot3\sqrt[3]{24}}\)
Tương tự: \(\sqrt{2y^2+3yz+4z^2}\ge\sqrt{yz\cdot3\sqrt[3]{24}}\); \(\sqrt{2z^2+3zx+4x^2}\ge\sqrt{zx\cdot3\sqrt[3]{24}}\)
Cộng theo vế 3 BĐT vừa tìm, ta được:
\(P\ge\sqrt{3\sqrt[3]{24}}\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)=\sqrt{3\sqrt[3]{24}}=\sqrt[6]{648}\)
\(P=\sqrt{\frac{1}{36}\left(11a+7b\right)^2+\frac{59\left(a-b\right)^2}{36}}+\sqrt{\frac{1}{36}\left(7a+11b\right)+\frac{59\left(a-b\right)^2}{36}}\)
\(=\sqrt{\frac{1}{16}\left(3a+5b\right)^2+\frac{5\left(a-b\right)^2}{16}}+\sqrt{\frac{1}{16}\left(5a+3b\right)^2+\frac{5\left(a-b\right)^2}{16}}\)
\(\ge\frac{1}{6}\left(11a+7b\right)+\frac{1}{6}\left(7a+11b\right)+\frac{1}{4}\left(3a+5b\right)+\frac{1}{4}\left(5a+3b\right)\)
\(=5\left(a+b\right)=5.2016=10080\)