\(x+y=1\Rightarrow y=1-x\)

\(P=x^3+\left(1-x\right)^3+x\left(1-x\right)\)

\(P=2x^2-2x+1=\dfrac{1}{2}\left(2x-1\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}\)

\(P_{min}=\dfrac{1}{2}\) khi \(x=y=\dfrac{1}{2}\)

Ta có: \(2x^2 + \frac{y^2}{4} + \frac{1}{x^2} = 4\)

=>\(\frac{y^2}{4} = 4 - 2x^2 - \frac{1}{x^2}\)

=>\(4-2x^2-\frac{1}{x^2}>0\iff2x^2+\frac{1}{x^2}<4\)

=>\(2x^4 - 4x^2 + 1 < 0\)

=>\(\frac{2 - \sqrt{2}}{2} < x^2 < \frac{2 + \sqrt{2}}{2}\)

\(y^2 = 4\left(4 - 2x^2 - \frac{1}{x^2}\right) = 16 - 8x^2 - \frac{4}{x^2}\)

=>\((xy)^2 = x^2 y^2 = x^2 \left(16 - 8x^2 - \frac{4}{x^2}\right) = 16x^2 - 8x^4 - 4\)

Đặt \(t=x^2\left(\frac{2-\sqrt2}{2}

Xét hàm số \(f(t) = -8t^2 + 16t - 4\)

Hoành độ đỉnh là \(t = -\frac{16}{2 \cdot (-8)} = 1\)

\(f(1) = -8(1)^2 + 16(1) - 4 = 4\)

=>0<=(xy)^2<=4

=>-2<=xy<=2

=>\(2018-2\le2018+xy\le2018+2\)

=>2016<=A<=2020

=>A min=2016 khi xy=-2; A max=2020 khi xy=2

Ta có \(x^2+y^2+xy+x=y-1\)

\(\Leftrightarrow2x^2+2y^2+2xy+2x-2y+2=0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(x+1\right)^2+\left(y-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x+1=0\\y-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)

\(\Rightarrow B=\left(-1+1-1\right)^{2023}\) \(=\left(-1\right)^{2023}\) \(=-1\)

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