Question

\(\left\{{}\begin{matrix}2xy=x+y+1\\\frac{x^2}{\left(y+1\right)^2}+\frac{y^2}{\left(x+1\right)^2}=\frac{10}{9}\end{matrix}\right.\)

Answer 1

Đk: x, y\(\ne\) -1

Xét x, y bằng 0 => hpt vô nghiệm

Đặt \(\frac{x}{y+1}=a,\frac{y}{x+1}=b\)

=> \(ab=\frac{xy}{\left(y+1\right)\left(x+1\right)}=\frac{xy}{xy+x+y+1}=\frac{xy}{xy+2xy}=\frac{xy}{3xy}=\frac{1}{3}\)

<=> \(a=\frac{1}{3b}\)

Có \(a^2+b^2=\frac{10}{9}\)<=> \(\left(\frac{1}{3b}\right)^2+b^2=\frac{10}{9}\)

<=> \(9+81b^4=90b^{^2}\) <=> \(9b^4-10b^2+1=0\)

<=> \(\left(b^2-1\right)\left(9b^2-1\right)=0\) <=> \(\left(b-1\right)\left(b+1\right)\left(3b-1\right)\left(3b+1\right)=0\) <=> \(\left[{}\begin{matrix}b=1\\b=-1\\b=\frac{1}{3}\\b=-\frac{1}{3}\end{matrix}\right.\)

<=>\(\left[{}\begin{matrix}a=\frac{1}{3}\\a=-\frac{1}{3}\\a=1\\a=-1\end{matrix}\right.\)

Tại \(\left(a,b\right)=\left(\frac{1}{3},1\right)\) => \(\left(x;y\right)=\left(1;2\right)\)

Tại \(\left(a,b\right)=\left(-\frac{1}{3},-1\right)\) => \(\left(x;y\right)\in\varnothing\)

Tại \(\left(a,b\right)=\left(1,\frac{1}{3}\right)\)=> \(\left(x;y\right)=\left(2;1\right)\)

Tại \(\left(a,b\right)=\left(-1,-\frac{1}{3}\right)\) =>\(\left(x,y\right)\in\varnothing\)

Vậy hpt có 2 tập nghiệm duy nhất (1,2) , (2,1)