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Đáp án: $(x,y)\in\{(1,1),(\dfrac23, -\dfrac13)\}$

Giải thích các bước giải:

Ta có:

$\begin{cases} 4xy+4(x^2+y^2)+\dfrac{3}{(x+y)^2}=\dfrac{85}{3}\\ 2x+\dfrac{1}{x+y}=\dfrac{13}{3}\end{cases}$

$\to \begin{cases} (x-y)^2+3(x+y)^2+\dfrac{3}{(x+y)^2}=\dfrac{85}{3}\\ (x-y)+(x+y)+\dfrac{1}{x+y}=\dfrac{13}{3}\end{cases}$

$\to \begin{cases} (x-y)^2+3((x+y)^2+\dfrac{1}{(x+y)^2})=\dfrac{85}{3}\\ (x-y)+((x+y)+\dfrac{1}{x+y})=\dfrac{13}{3}\end{cases}$

$\to \begin{cases} (x-y)^2+3((x+y)+\dfrac{1}{x+y})^2=\dfrac{103}{3}\\ (x-y)+((x+y)+\dfrac{1}{x+y})=\dfrac{13}{3}\end{cases}$

$\to \begin{cases} (x-y)^2+3(x+y+\dfrac{1}{x+y})^2=\dfrac{103}{3}\\ (x-y)+((x+y)+\dfrac{1}{x+y})=\dfrac{13}{3}\end{cases}$

Đặt $x-y=a, (x+y)+\dfrac{1}{x+y}=b$

$\to \begin{cases} a^2+3b^2=\dfrac{103}{3}\\ a+b=\dfrac{13}{3}\end{cases}$

$\to \begin{cases} a^2+3(\dfrac{13}{3}-a)^2=\dfrac{103}{3}\\ b=\dfrac{13}{3}-a\end{cases}$

$\to \begin{cases} a\in\{1,\dfrac{11}{2}\}\\ b=\dfrac{13}{3}-a\end{cases}$

$\to (a,b)\in\{(1,\dfrac{10}{3}), (\dfrac{11}{2}, -\dfrac76)\}$

Nếu $(a,b)=(1,\dfrac{10}{3})$

$\to\begin{cases}x-y=1\\ x+y+\dfrac{1}{x+y}=\dfrac{10}{3}\end{cases}$ 

$\to\begin{cases}x=y+1\\ (y+1)+y+\dfrac{1}{(y+1)+y}=\dfrac{10}{3}\end{cases}$ 

$\to (x,y)\in\{(1,1),(\dfrac23, -\dfrac13)\}$

Nếu $(a,b)=(\dfrac{11}{2}, -\dfrac76)$

$\to\begin{cases}x-y=\dfrac{11}{2}\\ x+y+\dfrac{1}{x+y}=\dfrac{-7}{6}\end{cases}$ 

$\to\begin{cases}x=y+\dfrac{11}{2}\\ (y+\dfrac{11}{2})+y+\dfrac{1}{y+\dfrac{11}{2}+y}=\dfrac{-7}{6}\end{cases}$ 

$\to$Vô nghiệm