a: Thay x=-2 vào phương trình, ta được:
\(\left(m-1\right)\cdot\left(-2\right)^2-2\left(m+2\right)\cdot\left(-2\right)+m-1=0\)
=>4(m-1)+4(m+2)+m-1=0
=>5m-5+4m+8=0
=>9m+3=0
=>9m=-3
=>\(m=-\frac39=-\frac13\)
b: TH1: m=1
Phương trình sẽ trở thành:
\(\left(1-1\right)\cdot x^2-2\left(1+2\right)x+1-1=0\)
=>-6x=0
=>x=0
=>Loại
TH2: m<>1
\(\Delta=\left\lbrack-2\left(m+2\right)\right\rbrack^2-4\left(m-1\right)\left(m-1\right)\)
\(=4\left(m+2\right)^2-4\left(m-1\right)^2=4\left(m^2+4m+4-m^2+2m-1\right)=4\left(6m+3\right)\)
Để phương trình có hai nghiệm phân biệt thì 6m+3>0
=>6m>-3
=>m>-1/2
Theo Vi-et, ta có: \(x_1+x_2=-\frac{b}{a}=\frac{2\left(m+2\right)}{m-1};x_1x_2=\frac{c}{a}=\frac{m-1}{m-1}=1\)
\(\frac{x_1}{x_2}+\frac{x_2}{x_1}=62\)
=>\(\frac{x_1^2+x_2^2}{x_1x_2}=62\)
=>\(x_1^2+x_2^2=62\)
=>\(\left(x_1+x_2\right)^2-2x_1x_2=62\)
=>\(\left\lbrack\frac{2\left(m+2\right)}{m-1}\right\rbrack^2-2\cdot1=62\)
=>\(4\left(\frac{m+2}{m-1}\right)^2=62+2=64\)
=>\(\left(\frac{m+2}{m-1}\right)^2=16\)
=>\(\left[\begin{array}{l}\frac{m+2}{m-1}=4\\ \frac{m+2}{m-1}=-4\end{array}\right.\Rightarrow\left[\begin{array}{l}4\left(m-1\right)=m+2\\ -4\left(m-1\right)=m+2\end{array}\right.\)
=>\(\left[\begin{array}{l}4m-4-m-2=0\\ -4m+4-m-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}3m-6=0\\ -5m+2=0\end{array}\right.\)
=>\(\left[\begin{array}{l}m=2\left(nhận\right)\\ m=\frac25\left(nhận\right)\end{array}\right.\)
