Bài 2:
\(\left\{{}\begin{matrix}3x-y=2m-1\\x+2y=3m+2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}6x-2y=4m-2\\x+2y=3m+2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}7x=7m\\x+2y=3m+2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=m\\y=m+1\end{matrix}\right.\)
Ta có:
\(x^2+y^2=10\)
\(\Leftrightarrow m^2+\left(m+1\right)^2=10\)
\(\Leftrightarrow2m^2+2m-9=0\)
\(\Delta=b^2-4ac=2^2-4.2.\left(-9\right)=76\)
Vì \(\Delta=76>0\) ⇒ Phương trình có hai nghiệm phân biệt
\(m_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-2+\sqrt{76}}{2.2}=\dfrac{-1+\sqrt{19}}{2}\)
\(m_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-2-\sqrt{76}}{2.2}=\dfrac{-1+\sqrt{19}}{2}\)
Vậy hệ phương trình có nghiệm duy nhất \(\left(x;y\right)=\left(\dfrac{-1+\sqrt{19}}{2};\dfrac{-1-\sqrt{19}}{2}\right)\)
Bài 1:
\(\left\{{}\begin{matrix}3x+y=5-3m\\x+y=5m^2+4m\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=m^2-2m+2\\y=3m^2+3m-1\end{matrix}\right.\)
Ta có:
\(A=x+y\)
\(A=-m^2-2m+2+3m^2+3m-1\)
\(A=2m^2+m+1\)
\(A=2\left(m^2+2.\dfrac{1}{4}m+\dfrac{1}{16}\right)+\dfrac{7}{8}\ge0\)
Vậy GTNN của A bằng \(\dfrac{7}{8}\) khi m = \(\dfrac{-1}{4}\)
