1 ) Đặt \(x+y=S;xy=p\) , ta có :
\(\left\{{}\begin{matrix}S+p=7\\S^2-p=13\end{matrix}\right.\) \(\Rightarrow S^2+S=20\Leftrightarrow\left(S-4\right)\left(S+5\right)=0\Leftrightarrow\left[{}\begin{matrix}S=4\\S=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}p=3\\p=12\end{matrix}\right.\)
TH 1 : \(S=4;p=3\) . Giải pt : \(x^2-4x+3=0\)
TH 2 : S \(=-5;p=12\) . Giải pt : \(x^2+5x+12=0\)
( tự giải nha )
2 ) Ta có HPT :
\(\left\{{}\begin{matrix}x+y=7\\x^3+y^3=133\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=49\\\left(x+y\right)\left(x^2-xy+y^2\right)=133\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x^2+2xy+y^2=49\\x^2-xy+y^2=19\end{matrix}\right.\)
\(\Rightarrow3xy=30\Leftrightarrow xy=10\)
\(\left\{{}\begin{matrix}x+y=7\\xy=10\end{matrix}\right.\) => pt : \(x^2-7x+10=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=5\\y=2\end{matrix}\right.\)
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