Bài 3:
\(\Delta=\left[-2\left(m-1\right)\right]^2-4\cdot1\cdot\left(m^2-4\right)\)
\(=\left(2m-2\right)^2-4\left(m^2-4\right)\)
\(=4m^2-8m+4-4m^2+16=-8m+20\)
Để phương trình có hai nghiệm thì \(\Delta>=0\)
=>-8m+20>=0
=>-8m>=-20
=>\(m< =\dfrac{5}{2}\)
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2m-2\\x_1x_2=\dfrac{c}{a}=m^2-4\end{matrix}\right.\)
\(Q=\left(x_1-3x_2\right)\left(x_2-3x_1\right)-5m^2-20m\)
\(=x_1x_2-3x_1^2-3x_2^2+9x_1x_2-5m^2-20m\)
\(=-3\left(x_1^2+x_2^2\right)+10x_1x_2-5m^2-20m\)
\(=-3\left(x_1+x_2\right)^2+16x_1x_2-5m^2-20m\)
\(=-3\left(2m-2\right)^2+16\left(m^2-4\right)-5m^2-20m\)
\(=-3\left(4m^2-8m+4\right)+16m^2-64-5m^2-20m\)
\(=-12m^2+24m-12+11m^2-20m-64\)
\(=-m^2+4m-76\)
\(=-\left(m^2-4m+76\right)\)
\(=-\left(m^2-4m+4+72\right)\)
\(=-\left(m-2\right)^2-72< =-72\forall m\)
Dấu '=' xảy ra khi m-2=0
=>m=2(nhận)
Bài 2:
a: \(\Delta=\left[-\left(2m-3\right)\right]^2-4\cdot1\cdot\left(m^2-3m\right)\)
\(=4m^2-12m+9-4m^2+12m=9>0\)
=>Phương trình luôn có hai nghiệm phân biệt
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2m-3\\x_1x_2=\dfrac{c}{a}=m^2-3m\end{matrix}\right.\)
\(\dfrac{1}{x_1-5}+\dfrac{1}{x_2-5}=-\dfrac{1}{2}\)
=>\(\dfrac{x_2-5+x_1-5}{\left(x_1-5\right)\left(x_2-5\right)}=\dfrac{-1}{2}\)
=>\(\dfrac{x_1+x_2-10}{x_1x_2-5\left(x_1+x_2\right)+25}=\dfrac{-1}{2}\)
=>\(\dfrac{2m-3-10}{m^2-3m-5\left(2m-3\right)+25}=\dfrac{-1}{2}\)
=>\(\dfrac{2m-13}{m^2-3m-10m+15+25}=\dfrac{-1}{2}\)
=>\(\dfrac{2m-13}{m^2-13m+40}=\dfrac{-1}{2}\)
=>\(-m^2+13m-40=4m-26\)
=>\(-m^2+13m-40-4m+26=0\)
=>\(-m^2+9m-14=0\)
=>(m-2)(m-7)=0
=>\(\left[{}\begin{matrix}m=2\\m=7\end{matrix}\right.\)
b: \(P=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2\)
\(=\left(2m-3\right)^2-2\left(m^2-3m\right)\)
\(=4m^2-12m+9-2m^2+6m\)
\(=2m^2-6m+9\)
\(=2\left(m^2-3m+\dfrac{9}{2}\right)\)
\(=2\left(m^2-3m+\dfrac{9}{4}+\dfrac{9}{4}\right)\)
\(=2\left(m-\dfrac{3}{2}\right)^2+\dfrac{9}{2}>=\dfrac{9}{2}\forall m\)
Dấu '=' xảy ra khi \(m-\dfrac{3}{2}=0\)
=>\(m=\dfrac{3}{2}\)
