\(\Delta=\left\lbrack-\left(2m+1\right)\right\rbrack^2-4\cdot1\cdot\left(-m^2-2\right)\)
\(=4m^2+4m+1+4m^2+8=8m^2+4m+9\)
\(=8\left(m^2+\frac12m+\frac98\right)\)
\(=8\left(m^2+\frac12m+\frac{1}{16}+\frac98-\frac{1}{16}\right)\)
\(=8\left(m+\frac14\right)^2+8\cdot9-8\cdot\frac{1}{16}=8\left(m+\frac14\right)^2+72-\frac12=8\left(m+\frac14\right)^2+71,5\ge71,5>0\forall m\)
=>Phương trình luôn có hai nghiệm phân biệt
Theo Vi-et, ta có: \(\begin{cases}x_1+x_2=-\frac{b}{a}=2m+1\\ x_1x_2=\frac{c}{a}=-m^2-2\end{cases}\)
\(A=\frac{x_1+x_2}{x_1x_2}\)
\(=\frac{2m+1}{-m^2-2}=-\frac{2m+1}{m^2+2}\)
Đặt A=k
=>\(-2m-1=k\left(m^2+2\right)\)
=>\(\operatorname{km}^2+2k+2m+1=0\)
=>\(\operatorname{km}^2+2m+2k+1=0\) (1)
\(\Delta=2^2-4k\left(2k+1\right)=4-8k^2-4k=-4\left(2k^2+k-1\right)\)
=-4(k+1)(2k-1)
Để (1) có nghiệm thì Δ>=0
=>-4(k+1)(2k-1)>=0
=>(k+1)(2k-1)<=0
=>\(-1\le k\le\frac12\)
=>\(-1\le A\le\frac12\)
=>\(A_{\min}=-1\) và \(A_{max}=\frac12\)
\(A_{\min}=-1\)
=>\(-\frac{2m+1}{m^2+2}=-1\)
=>\(m^2+2=2m+1\)
=>\(m^2-2m+1=0\)
=>\(\left(m-1\right)^2=0\)
=>m-1=0
=>m=1
\(A_{\max}=\frac12\)
=>\(\frac{-2m-1}{m^2+2}=\frac12\)
=>\(m^2+2=2\left(-2m-1\right)=-4m-2\)
=>\(m^2+4m+4=0\)
=>\(\left(m+2\right)^2=0\)
=>m+2=0
=>m=-2
