Thay y=1/2 vào pt ta dc
\(\dfrac{1}{2}\left(\left(\dfrac{1}{2}\right)^2+\dfrac{7}{4}\right)-2.\dfrac{1}{2}\left(m-1\right)=2m^2-8\)
<=>\(\dfrac{1}{2}\left(\dfrac{1}{4}+\dfrac{7}{4}\right)-m+1=2m^2-8\)
<=>\(\dfrac{1}{2}.2-m+1=2m^2-8\)
<=>\(2m^2+m-10=0\)
<=>(m-2)(2m+5)=0
<=>\(\left[{}\begin{matrix}m=2\\m=\dfrac{-5}{2}\end{matrix}\right.\)
Câu 1
Thay \(y=\dfrac{1}{2}\) vào phương trình đã cho ta có:
\(\dfrac{1}{2}\left[\left(\dfrac{1}{2}\right)^2+\dfrac{7}{4}\right]-2.\dfrac{1}{2}\left(m-1\right)=2m^2-8\)
\(\Rightarrow1-m+1=2m^2-8\)
\(\Rightarrow2m^2+m-10=0\)
\(\Rightarrow\left(2m+5\right)\left(m-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2m+5=0\\m-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}m=-\dfrac{5}{2}\\m=2\end{matrix}\right.\)
Câu 2:
Thay \(x=2\) vào phương trình đã cho ta có:
\(\left(2^2+1\right)=m\left(2+3\right)\)
\(\Rightarrow5m=5\)
\(\Rightarrow m=1\)
