Question

\(\left\{{}\begin{matrix}\left(x+y\right)\left(x^2+y^2\right)=15\\\left(x-y\right)\left(x^2-y^2\right)=3\end{matrix}\right.\)

Answer 1

\(\left\{{}\begin{matrix}\left(x+y\right)\left(x^2+y^2\right)=15\\\left(x-y\right)\left(x^2-y^2\right)=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)\left(x^2+y^2\right)=15\left(1\right)\\\left(x-y\right)^2\left(x+y\right)=3\left(2\right)\end{matrix}\right.\)

Lấy (1) : (2) ta có :

\(x^2+y^2=5.\left(x-y\right)^2\Leftrightarrow x^2+y^2=5x^2-10xy+5y^2\)

\(\Leftrightarrow4\left(x^2+y^2\right)=10xy\Leftrightarrow x^2+y^2=\dfrac{10}{4}xy\left(3\right)\)

thay (3) vào (1) ta có :

(x+y) .xy=6

=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y=2\\xy=3\end{matrix}\right.\\\left\{{}\begin{matrix}x+y=1\\xy=6\end{matrix}\right.\end{matrix}\right.\)

còn lại bn tự giải