Question

tìm giá trị nguyên của m để hệ phương trình \(\left\{{}\begin{matrix}3x-y=2m-1\\x+2y=3m+2\end{matrix}\right.\) có nghiệm(x,y) thỏa mãn :\(x^2\) + \(2y^2\)= 9

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}6x-2y=4m-2\\x+2y=3m+2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7x=7m\\y=3x-2m+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=m\\y=m+1\end{matrix}\right.\)

\(x^2+2y^2=9\Leftrightarrow m^2+2\left(m+1\right)^2=9\)

\(\Leftrightarrow3m^2+4m-7=0\)

\(\Rightarrow\left[{}\begin{matrix}m=1\\m=-\dfrac{7}{3}\end{matrix}\right.\)

Answer 2

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

\(\Leftrightarrow\left\{{}\begin{matrix}6x-2y=4m-2\\x+2y=3m+2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7x=4m-2+3m+2\\x+2y=3m+2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7x=7m\\x+2y=3m+2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=m\\2y=2m+2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=m\\y=m+1\end{matrix}\right.\)

Mà: \(x^2+2y^2=9\)

\(\Leftrightarrow m^2+2\left(m+1\right)^2=9\)

\(\Leftrightarrow m^2+2\left(m^2+2m+1\right)=9\)

\(\Leftrightarrow m^2+2m^2+4m+2-9=0\)

\(\Leftrightarrow3m^2+4m-7=0\)

\(\Delta=4^2-4\cdot3\cdot-7=100>0\)

Pt có 2 nghiệm phân biệt:

\(m_1=\dfrac{-4+\sqrt{100}}{2\cdot3}=1\)

\(m_2=\dfrac{-4-\sqrt{100}}{6}=\dfrac{-7}{3}\)