Bài 20:
\(\Delta=\left\lbrack-\left(m-3\right)\right\rbrack^2-4\cdot1\cdot\left(-m\right)\)
\(=m^2-6m+9+4m=m^2-2m+9\)
\(=m^2-2m+1+8=\left(m-1\right)^2+8\ge8>0\forall m\)
=>Phương trình luôn có hai nghiệm phân biệt
Bài 21:
\(\Delta=\left\lbrack-2\left(m-2\right)\right\rbrack^2-4\cdot1\cdot\left(-2m+3\right)\)
\(=4\left(m^2-4m+4\right)+4\left(2m-3\right)\)
\(=4\left(m^2-4m+4+2m-3\right)=4\left(m^2-2m+1\right)=4\left(m-1\right)^2\ge0\forall m\)
=>Phương trình luôn có hai nghiệm
Theo Vi-et, ta có:
\(\begin{cases}x_1+x_2=-\frac{b}{a}=2\left(m-2\right)\\ x_1x_2=\frac{c}{a}=-2m+3\end{cases}\)
\(\left|x_1-x_2\right|=5\)
=>\(\left(x_1-x_2\right)^2=5^2=25\)
=>\(\left(x_1+x_2\right)^2-4x_1x_2=25\)
=>\(\left(2m-4\right)^2-4\left(-2m+3\right)=25\)
=>\(4m^2-16m+16+8m-12=25\)
=>\(4m^2-8m+4=25\)
=>\(\left(2m-2\right)^2=\) 25
=>\(\left[\begin{array}{l}2m-2=5\\ 2m-2=-5\end{array}\right.\Rightarrow\left[\begin{array}{l}2m=7\\ 2m=-5+2=-3\end{array}\right.\Rightarrow\left[\begin{array}{l}m=\frac72\\ m=-\frac32\end{array}\right.\)
