đk: \(x,y\ne-2\)

\(hpt\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x+2}+\dfrac{y}{x+2}=1\\\left(\dfrac{x}{y+2}\right)^2+\left(\dfrac{y}{x+2}\right)^2=1\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}a=\dfrac{x}{y+2}\\b=\dfrac{y}{x+2}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{y+2}+\dfrac{y}{x+2}=1\\\left(\dfrac{x}{y+2}\right)^2+\left(\dfrac{y}{x+2}\right)^2=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=1\\a^2+b^2=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=1-a\\a^2+\left(1-a\right)^2=1\end{matrix}\right.\)

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}a=0\\b=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}a=1\\b=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\end{matrix}\right.\)

- đường thẳng (d) tiếp xúc với (P) có hệ số góc k=y'=-x

- đường thẳng (d) đi qua A(-2;-2) => k=-x_A=2

==> pt đường thẳng (d) là : y=2(x+2)-2 <=> y=2x+2

\(\left\{{}\begin{matrix}xy+3y^2+x=3\left(1\right)\\x^2+xy-2y^2\left(2\right)\end{matrix}\right.\)

\(pt\left(2\right)\Leftrightarrow\left(x^2-y^2\right)+y\left(x-y\right)=0\Leftrightarrow\left(x-y\right)\left(x+2y\right)=0\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-2y\end{matrix}\right.\)

+) Với x=y, thay vào pt (1) ta có: \(4x^2+x-3=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{3}{4}\end{matrix}\right.\)

=> \(x=y=-1;x=y=\dfrac{3}{4}\)

+) Với \(x=-2y\), thay vào pt(1) ta có: \(y^2-2y-3=0\Leftrightarrow\left[{}\begin{matrix}y=-1\Rightarrow x=2\\y=3\Rightarrow x=-6\end{matrix}\right.\)

Vậy hpt có 4 nghiệm: \(\left(x;y\right)\in\left\{\left(-1;-1\right),\left(\dfrac{3}{4};\dfrac{3}{4}\right),\left(2;-1\right),\left(-6;3\right)\right\}\)

Vì \(x=\sqrt{\dfrac{1}{2\sqrt{3}-2}-\dfrac{3}{2\sqrt{3}+2}}=\dfrac{\sqrt{3}-1}{2}\)

\(\Rightarrow x=\dfrac{\sqrt{3}-1}{2}\) là nghiệm của đa thức \(2x^2+2x-1\)

\(\Rightarrow B=\dfrac{2x^{2017}\left(2x^2+2x-1\right)+2x+1}{\left(2x^2+2x-1\right)+x+1}=\dfrac{2x+1}{x+1}=3-\sqrt{3}\)

Ta có: \(P=ab+\dfrac{4}{ab}+4\ge2\sqrt{ab.\dfrac{4}{ab}+4}=8\)

Dấu '=' xảy ra <=> \(\left\{{}\begin{matrix}ab=2\\1\le a,b\le2\end{matrix}\right.\)

Lại có: \(1\le a\le2,1\le b\le2\)

\(\Rightarrow1\le ab\le4\Leftrightarrow\left(ab-1\right)\left(ab-4\right)\le0\Leftrightarrow\left(ab\right)^2\le5ab-4\)

\(\Rightarrow P=\dfrac{\left(ab\right)^2+4ab+4}{ab}\le\dfrac{5ab-4+4ab+4}{ab}=9\)

Dấu '=' xảy ra <=> \(\left[{}\begin{matrix}ab=1\\ab=4\end{matrix}\right.\) và \(1\le a,b\le2\) \(\Leftrightarrow\left[{}\begin{matrix}a=b=2\\a=b=1\end{matrix}\right.\)

Vậy \(Min_P=8\Leftrightarrow ab=2;1\le a,b\le2\)

\(Max_P=9\Leftrightarrow\left[{}\begin{matrix}a=b=1\\a=b=2\end{matrix}\right.\)

D