$\left \{ {{x²+y²=1} \atop {-x²-5xy+2y²=3}} \right.$ 13/07/2021 Bởi Skylar $\left \{ {{x²+y²=1} \atop {-x²-5xy+2y²=3}} \right.$
\begin{cases} x^2+y^2=1\\-x^2+5xy+2y^2=3 \end{cases} `⇔-x^2+5xy+2y^2=3x^2+3y^2` `⇔4x^2-5xy+y^2=0` `⇔(x-y)(4x-y)=0` `⇔`\(\left[ \begin{array}{l}x-y=0\\4x-y=0\end{array} \right.\) `⇔`\(\left[ \begin{array}{l}x=y\\4x=y\end{array} \right.\) `⇔`\(\left[ \begin{array}{l}2x^2=1\\17x^2=1\end{array} \right.\) `⇔`\(\left[ \begin{array}{l}x=1/(±√2)\\x=1/(±√17)\end{array} \right.\) Bình luận
\begin{cases} x^2+y^2=1\\-x^2+5xy+2y^2=3 \end{cases}
`⇔-x^2+5xy+2y^2=3x^2+3y^2`
`⇔4x^2-5xy+y^2=0`
`⇔(x-y)(4x-y)=0`
`⇔`\(\left[ \begin{array}{l}x-y=0\\4x-y=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=y\\4x=y\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}2x^2=1\\17x^2=1\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=1/(±√2)\\x=1/(±√17)\end{array} \right.\)