Giải hệ phương trình x^2 +1+y(y+x)=4y và (x^2+1)(y+x-2)=y có nghiệm (x;y

* Hướng dẫn giải

+) Xét y = 0 hệ phương trình đã cho trở thành $\left\{\begin{array}{l}{x}^2+1=0\\ \left(x^2+1\right)\left(x-2\right)=0\end{array}\right.$(vô lý)

+) Xét y ≠ 0 chia các vế của từng phương trình cho y ta được:

$\left\{\begin{array}{l}\frac{x^2+1}{y}+y+x=4\\ \frac{x^2+1}{y}\left(y+x-2\right)=1\end{array}\right.$

Đặt $\left\{\begin{array}{l}\frac{x^2+1}{y}=a\\ y+x-2=b\end{array}\right.$

$$\begin{gathered} \Rightarrow \left\{\begin{array}{l}a+b=2\\ ab=1\end{array}\right.\iff \left\{\begin{array}{l}a=2-b\\ a(2-a)=1\end{array}\right. \\ \iff \left\{\begin{array}{l}b=2-a\\ a^2-2a+1=0\end{array}\right.\iff \left\{\begin{array}{l}b=2-a\\ {\left(a-1\right)}^2=0\end{array}\right. \\ \iff a=b=1\iff \left\{\begin{array}{l}\frac{x^2+1}{y}=1\\ y+x-2=1\end{array}\right. \\ \iff \left\{\begin{array}{l}y=x^2+1\\ x+y=3\end{array}\right.\iff \left\{\begin{array}{l}y=x^2+1\\ x+x^2+1=3\end{array}\right. \\ \iff \left\{\begin{array}{l}y=x^2+1\\ x^2+x-2=0\end{array}\right.\iff \left\{\begin{array}{l}y=x^2+1\\ \left(x-1\right)\left(x+2\right)=0\end{array}\right. \\ \iff \left\{\begin{array}{l}y=x^2+1\\ \left[\begin{array}{l}x=1\\ x=-2\end{array}\right.\end{array}\right.\iff \left[\begin{array}{l}\left\{\begin{array}{l}x=1\\ y=2\end{array}\right.\left(tm\right)\\ \left\{\begin{array}{l}x=-2\\ y=5\end{array}\right.\left(tm\right)\end{array}\right. \end{gathered}$$

Đáp án:D