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a: (2x-3)^2>=0
=>-(2x-3)^2<=0
=>D<=-3
Dấu = xảy ra khi x=3/2
b: (2x-5)^2>=0
(y+1/2)^2>=0
=>(2x-5)^2+(y+1/2)^2>=0
=>D>=2022
Dấu = xảy ra khi x=5/2 và y=-1/2
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\(B=\left(x-2y\right)^2+y^2+2x+6y+2046=\left[\left(x-2y\right)^2+2\left(x-2y\right)+1\right]+\left(y^2+10y+25\right)+2020=\left(x-2y+1\right)^2+\left(y+5\right)^2+2020\ge2020\)
\(minB=2020\Leftrightarrow\)\(\left\{{}\begin{matrix}x=-11\\y=-5\end{matrix}\right.\)
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c: Ta có: \(\left(x+1\right)^2\ge0\forall x\)
\(\left(y-\dfrac{1}{3}\right)^2\ge0\forall y\)
Do đó: \(\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2\ge0\forall x,y\)
\(\Leftrightarrow\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2-10\ge-10\forall x,y\)
Dấu '=' xảy ra khi x=-1 và \(y=\dfrac{1}{3}\)
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