Đề 21-22
a: \(\Delta=\left(2m-1\right)^2-4m\left(m-2\right)\)
\(=4m^2-4m+1-4m^2+8m=4m+1\)
Để phương trình có nghiệm thì \(\left\{{}\begin{matrix}m-2\ne0\\4m+1>=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< >2\\m>=-\dfrac{1}{4}\end{matrix}\right.\)
b: Áp dụng hệ thức Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-2m+1}{m-2}\\x_1\cdot x_2=\dfrac{m}{m-2}\end{matrix}\right.\)
Ta có: \(\left(x_1+x_2\right)^2+3x_1x_2=2\)
\(\Leftrightarrow\left(\dfrac{-2m+1}{m-2}\right)^2+\dfrac{3m}{m-2}=\dfrac{2m-4}{m-2}\)
\(\Leftrightarrow\left(-2m+1\right)^2+3m\left(m-2\right)=2\left(m-2\right)^2\)
\(\Leftrightarrow4m^2-4m+1+3m^2-6m-2m^2+8m-8=0\)
\(\Leftrightarrow5m^2+2m-7=0\)
\(\Leftrightarrow\left(5m+7\right)\left(m-1\right)=0\)
hay m=1
