\(\)đặt x+y=a , xy=b
=> \(\left\{{}\begin{matrix}a\left(1+\frac{1}{b}\right)=5\\\left(a^2-2b\right)\left(1+\frac{1}{b^2}\right)=49\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}a=\frac{5}{1+\frac{1}{b}}\\\left(\left(\frac{5}{1+\frac{1}{b}}\right)^2-2b\right)\left(1+\frac{1}{b^2}\right)=49\end{matrix}\right.\)
=> giải đc b2+7b+1=0 => b => a => x,y
số xấu vler , chả biết sếp nào nghĩ ra bài này nữa
ĐKXĐ: ...
\(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=5\\x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=49\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{1}{x}+y+\frac{1}{y}=5\\\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2=53\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+\frac{1}{x}=a\\y+\frac{1}{y}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=5\\a^2+b^2=53\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\ab=-14\end{matrix}\right.\) \(\Rightarrow a;b\) là nghiệm của \(t^2-5t-14=0\Rightarrow\left[{}\begin{matrix}t=7\\t=-2\end{matrix}\right.\)
\(\Rightarrow\left(a;b\right)=\left(-2;7\right);\left(7;-2\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x+\frac{1}{x}=-2\\y+\frac{1}{y}=7\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x+\frac{1}{x}=7\\y+\frac{1}{y}=-2\end{matrix}\right.\)
