Ta có: \(x+y=5\Rightarrow x=5-y\)
Thay vào ta được:
\(xy=\left(5-y\right).y=6\)
\(\Rightarrow5y-y^2=6\)
\(\Leftrightarrow y^2-5y+6=0\)
\(\Leftrightarrow\left(y-2\right)\left(y-3\right)=6\)
\(\Leftrightarrow\left[{}\begin{matrix}y=2\\y=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
Suy ra:
\(\left[{}\begin{matrix}x^2+y^2=3^2+2^2=13\\x^2+y^2=2^2+3^2=13\end{matrix}\right.\)
Vậy: \(x^2+y^2=13\)
giải:
Ta có: \(x+y=5\)
\(\Leftrightarrow\left(x+y\right)^2=5^2\)
\(\Leftrightarrow x^2+2xy+y^2=25\)
\(\Leftrightarrow x^2+2\left(-6\right)+y^2=25\)
\(\Leftrightarrow x^2-12+y^2=25\)
\(\Leftrightarrow x^2+y^2=25+12=37\)
Vậy ...
\(x+y=5\Rightarrow\left(x+y\right)^2=5^2=25\)
\(\Rightarrow x^2+2xy+y^2=25\)
\(\Rightarrow x^2+y^2+2.\left(-6\right)=25\)
\(\Rightarrow x^2+y^2+\left(-12\right)=25\)
\(\Rightarrow x^2+y^2=37\)
Ta có:\(x+y=5\)
\(\Rightarrow\left(x+y\right)^2=5^2=25\)
\(\Leftrightarrow x^2+2xy+y^2=25\)
\(x^2+2.\left(-6\right)+y^2=25\)
\(x^2-12+y^2=25\)
\(x^2+y^2=25+12=37\)
Vậy: \(x^2+y^2=37\)
