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\(P=\left(2x+\frac{1}{x}\right)^2+\left(2y+\frac{1}{y}\right)^2\)
Ta có: \(2x+\frac{1}{x}\ge2\sqrt{2x+\frac{1}{x}}=2\sqrt{2}\)
\(\Rightarrow\left(2x+\frac{1}{x}\right)^2\ge8\)
\(\Rightarrow\left(2y+\frac{1}{y}\right)^2\ge8\)
Dấu \("="\) xảy ra \(\Leftrightarrow x=y=\pm\frac{1}{2}\)
Vậy \(P_{min}=16\Leftrightarrow x=y=\pm\frac{1}{2}\)
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\(B=4x^2+5y^2-4xy+3x-y\)
\(\Leftrightarrow\left(4x^2-4xy+3x\right)+5y^2-y\)
\(\Leftrightarrow\left[4x^2-4x\left(y-\dfrac{3}{4}\right)+\left(y-\dfrac{3}{4}\right)^2\right]+5y^2-y-y^2+\dfrac{3}{2}y-\dfrac{9}{16}\)\(\Leftrightarrow\left(2x-y+\dfrac{3}{4}\right)^2+\left(4y^2-\dfrac{1}{2}y+\dfrac{1}{64}\right)-\dfrac{37}{64}\)
\(\Leftrightarrow\left(2x-y+\dfrac{3}{4}\right)^2+\left(2y-\dfrac{1}{8}\right)^2-\dfrac{37}{64}\ge\dfrac{-37}{64}\)
Vậy Min B = \(\dfrac{-37}{64}\) khi \(\left[{}\begin{matrix}\left(2x-y+\dfrac{3}{4}\right)^2=0\\\left(2y-\dfrac{1}{8}\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-y+\dfrac{3}{4}=0\\2y-\dfrac{1}{8}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-y+\dfrac{3}{4}=0\\2y=\dfrac{1}{8}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-\dfrac{1}{16}+\dfrac{3}{4}=0\\y=\dfrac{1}{16}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-11}{32}\\y=\dfrac{1}{16}\end{matrix}\right.\)
\(C=9y^2+2x^2-6y-6xy+5x-1\)
\(=\left(9y^2+6y-6xy\right)+2x^2+5x-1\)
\(=\left[9y^2+6y\left(1-x\right)+\left(1-x\right)^2\right]+2x^2+5x-1-1+2x-x^2\)\(=\left(3y-x+1\right)^2+\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{17}{4}\)
\(=\left(3y-x+1\right)^2+\left(x+\dfrac{3}{2}\right)^2-\dfrac{17}{4}\)
Vậy Min C = \(\dfrac{-17}{4}\) khi \(\left[{}\begin{matrix}\left(3y-x+1\right)^2=0\\\left(x+\dfrac{3}{2}\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3y-x+1=0\\x+\dfrac{3}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3y-\left(\dfrac{-3}{2}\right)+1=0\\x=\dfrac{-3}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=\dfrac{-5}{6}\\x=\dfrac{-3}{2}\end{matrix}\right.\)
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ta có : \(A=x^2-2x+y^2-4y+6=\left(x^2-2x+1\right)+\left(y^2-4y+4\right)+1\)
\(A=\left(x-1\right)^2+\left(y-2\right)^2+1\)
ta có : \(\left(x-1\right)^2\ge0\) với mọi \(x\) và \(\left(y-2\right)^2\ge0\) với mọi \(y\)
\(\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2+1\ge1\) với mọi \(x;y\)
\(\Rightarrow\) GTNN của \(A\) là 1 khi \(\left\{{}\begin{matrix}x-1=0\\y-2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
vậy giá trị nhỏ nhất của \(A\) là 1 khi \(x=1;y=2\)
A = \(x^2-2x+y^2-4y+6=x^2-2x+1+y^2-4y+4+1=\left(x-1\right)^2+\left(y-2\right)^2+1\ge1\)
Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}x-1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
Vậy GTNN của A là 1 khi x = 1 và y = 2
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Đặt
\(A=\left(2x-1\right)^2+\left(x+2\right)^2\)
=> \(A=4x^2-4x+1+x^2+4x+4\)
=> \(A=5x^2+5\)
=> \(Min_A=5\Leftrightarrow x=0\)
\(C=2x^2+y^2-2x\left(y-1\right)+3\Leftrightarrow2x^2+y^2-2xy+2x+3\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+2\Leftrightarrow\left(x-y\right)^2+\left(x+1\right)^2+2\ge2\)Vậy Min C = 2 khi \(\left[{}\begin{matrix}\left(x-y\right)^2=0\\\left(x+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-y=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=-1\\x=-1\end{matrix}\right.\)
\(C=2x^2+y^2-2x\left(y+1\right)+3\\ C=x^2-2xy+y^2+x^2-2x+1+2\\ C=\left(x-y\right)^2+\left(x-1\right)^2+2\)
vì: \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\\\left(x-1\right)^2\ge0\end{matrix}\right.\) nên \(C\ge2\)
dấu "=" xảy ra khi \(x-1=0\Rightarrow x=1\\ x-y=0\Leftrightarrow1-y=0\Rightarrow y=1\)
vậy GTNN của C là 2 tại x=y=1