\(\Delta=\left(-5\right)^2-4\cdot1\cdot\left(2m-1\right)\)
\(=25-8m+4\\ =29-8m\)
Để pt có nghiệm \(\Leftrightarrow\Delta\ge0\)
\(\Leftrightarrow29-8m\ge0\\ \Leftrightarrow-8m\ge-29\\ \Leftrightarrow m\le\dfrac{29}{8}\)
Với \(m\le\dfrac{29}{8}\) theo vi-ét ta có
\(\left\{{}\begin{matrix}x_1+x_2=5\\x_1\cdot x_2=2m-1\end{matrix}\right.\)
Có \(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}=\dfrac{19}{3}\)
\(\Leftrightarrow\dfrac{x_1^2+x^2_2}{x_1x_2}=\dfrac{19}{3}\)
\(\Leftrightarrow\dfrac{\left(x_1+x_2\right)^2-2x_1x_2}{x_1x_2}=\dfrac{19}{3}\)
\(\Leftrightarrow\dfrac{5^2-2\left(2m-1\right)}{2m-1}=\dfrac{19}{3}\) (đkxđ \(m\ne\dfrac{1}{2}\) )
\(\Leftrightarrow\dfrac{25-4m+2}{2m-1}=\dfrac{19}{3}\\ \Leftrightarrow\dfrac{27-4m}{2m-1}=\dfrac{19}{3}\\ \Leftrightarrow3\left(27-4m\right)=19\left(2m-1\right)\)
\(\Leftrightarrow81-12m=38m-19\\ \Leftrightarrow81+19=38m+12m\\ \Leftrightarrow100=50m\)
\(\Leftrightarrow m=2\) ( Thỏa mãn \(m\le\dfrac{29}{8};m\ne\dfrac{1}{2}\) )
Vậy......................................
-theo vi-ét ta có:
\(x_1x_2=\dfrac{c}{a}=2m-1\left(1\right)\)
\(x_1+x_2=\dfrac{-b}{a}=5\left(2\right)\)
- theo đề bài ta lại có:
\(\dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}=\dfrac{19}{3}\)
<=>\(\dfrac{x_1^2+x_2^2}{x_1x_2}=\dfrac{19}{3}\)
<=>\(\dfrac{x_1^2+2x_1x_2+x_2^2-2x_1x_2}{x_1x_2}=\dfrac{19}{3}\)
<=>\(\dfrac{\left(x_1+x_2\right)^2-2x_1x_2}{x_1x_2}=\dfrac{19}{3}\)(3)
-thay (1),(2) vào (3) ta được:
\(\dfrac{5^2-2\left(2m-1\right)}{2m-1}=\dfrac{19}{3}\)
=) m=2
vậy m=2
đề có yêu cầu x1,x2 là 2 nghiệm không ạ
