a. Em tự giải
b.
\(\Delta=\left(2m-1\right)^2-8\left(m-1\right)=4m^2-12m+9=\left(2m-3\right)^2\ge0;\forall m\)
\(\Rightarrow\) Pt luôn có nghiệm với mọi m
c.
Pt có 2 nghiệm pb khi \(2m-3\ne0\Rightarrow m\ne\dfrac{3}{2}\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-2m+1}{2}\\x_1x_2=\dfrac{m-1}{2}\end{matrix}\right.\)
\(3x_1+2x_2=1\)
\(\Leftrightarrow x_1+2\left(x_1+x_2\right)=1\)
\(\Leftrightarrow x_1-2m+1=1\)
\(\Rightarrow x_1=2m\)
Thế vào \(x_1+x_2=\dfrac{-2m+1}{2}\Rightarrow x_2=\dfrac{-2m+1}{2}-2m=\dfrac{-6m+1}{2}\)
Thế vào \(x_1x_2=\dfrac{m-1}{2}\)
\(\Rightarrow2m\left(\dfrac{-6m+1}{2}\right)=\dfrac{m-1}{2}\)
\(\Leftrightarrow-12m^2+m+1=0\Rightarrow\left[{}\begin{matrix}m=\dfrac{1}{3}\\m=-\dfrac{1}{4}\end{matrix}\right.\)
a)
Thay \(m=\dfrac{1}{2}\) vào (1) ta có:
\(2x^2+\left(2\cdot\dfrac{1}{2}-1\right)x+\dfrac{1}{2}-1=0\)
\(\Leftrightarrow2x^2-\dfrac{1}{2}=0\)
\(\Leftrightarrow2x^2=\dfrac{1}{2}\)
\(\Leftrightarrow x^2=\dfrac{1}{4}\)
\(\Leftrightarrow x=\pm\dfrac{1}{2}\)
b)
\(\Delta=\left(2m-1\right)^2-4\cdot2\cdot\left(m-1\right)=4m^2-4m+1-8m+8=4m^2-12m+9=\left(2m-3\right)^2\ge0\forall m\)
\(\Rightarrow\) PT luôn có nghiệm với mọi m
c) ĐK: m \(\ne\) \(\dfrac{3}{2}\)
Theo Vi-ét và đề bài ta có:\(\left\{{}\begin{matrix}x_1+x_2=-\left(2m-1\right)\left(1\right)\\x_1\cdot x_2=m-1\left(2\right)\\3x_1+2x_2=1\left(3\right)\end{matrix}\right.\)
Từ (1)(3) ta có: \(\left\{{}\begin{matrix}x_1+x_2=1-2m\\3x_1+2x_2=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x_1+2x_2=2-4m\\3x_1+2x_2=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-x_1=1-4m\\x_1+x_2=1-2m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1=4m-1\\4m-1+x_2=1-2m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1=4m-1\\x_2=2-6m\end{matrix}\right.\)
Thay vào (2) ta có:
\(\left(4m-1\right)\left(2-6m\right)=m-1\)
\(\Leftrightarrow-24m^2+14m-2=m-1\)
\(\Leftrightarrow-24m^2+13m-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m=\dfrac{13+\sqrt{73}}{48}\\m=\dfrac{13-\sqrt{73}}{48}\end{matrix}\right.\)(T/m)
