Ta có x+y=1
=> y=1-x
Khi đó
\(A=2x^2-\left(1-x\right)^2+x+\frac{1}{x}+1\)
\(=x^2+3x+\frac{1}{x}=\left(x-\frac{1}{2}\right)^2+\left(4x+\frac{1}{x}\right)-\frac{1}{4}\ge0+4-\frac{1}{4}=\frac{15}{4}\)
Vậy \(MinA=\frac{15}{4}\)khi \(x=\frac{1}{2},y=\frac{1}{2}\)
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