a,\(\hept{\begin{cases}x^2+y^2+\frac{2xy}{x+y}=1\\\sqrt{x+y}=x^2-y\end{cases}}\)

ĐK: \(x+y\ge0\)

\(\Leftrightarrow\hept{\begin{cases}\left(x+y\right)^2-2xy+\frac{2xy}{x+y}=1\left(1\right)\\\sqrt{x+y}=x^2-y\left(2\right)\end{cases}}\)

Đặt \(\hept{\begin{cases}x+y=a\\2xy=b\end{cases}\left(a\ge0\right)}\)

\(\left(1\right)\Leftrightarrow a^2-b+\frac{b}{a}=1\)

\(\Leftrightarrow a^3-ab-a+b=0\)

\(\Leftrightarrow\left(a-1\right)\left(a^2+a-b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=1\\a^2+a-b=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x+y=1\left(3\right)\\\left(x+y\right)^2+\left(x+y\right)-xy=0\left(4\right)\end{cases}}\)

Thay (3) vào (2)  ta được

\(x^2-y=1\Leftrightarrow y=x^2-1\)

\(\Rightarrow1-x=x^2-1\Leftrightarrow x^2+x-2=0\Leftrightarrow\orbr{\begin{cases}x=1\Rightarrow y=0\\x=-2\Rightarrow y=3\end{cases}}\)

Giải (4) 

Ta có \(\left(x+y\right)^2\ge4xy\Rightarrow\left(x+y\right)^2-xy>0\)

do đó (4) không xảy ra

Vậy..........

\(\hept{\begin{cases}a+b+c=4\\a^2+b^2+c^2=6\end{cases}}\)

\(b^2+c^2=6-a^2\Rightarrow\left(b+c\right)^2-2bc=6-a^2\)

\(\Rightarrow2bc=\frac{\left(b+c\right)^2-6+a^2}{2}\)

\(=\frac{\left(4-a\right)^2-6+a^2}{2}\left(Do:a+b+c=4\right)\)

\(=\frac{2a^2-8a+10}{2}=a^2-4a+5\)

\(\Rightarrow P=a^3+bc\left(b+c\right)=a^3+\left(a^2-4a+5\right)\left(4-a\right)\left(Do:a+b+c=4\right)\)

\(=a^3+4a^2-16a+20-a^3+4a^2-5a\)

\(=8a^2-21a+20\)

\(=8\left(a^2-2.\frac{21}{16}a+\frac{441}{256}\right)+\frac{199}{32}\)

\(=8\left(a-\frac{21}{16}\right)^2+\frac{119}{32}\)

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