\(\Delta=\left[-2\left(m+2\right)\right]^2-4\cdot1\cdot\left(-2m-5\right)\)
\(=\left(2m+4\right)^2+4\left(2m+5\right)\)
\(=4m^2+16m+16+8m+20\)
\(=4m^2+24m+36=4\left(m^2+6m+9\right)=4\left(m+3\right)^2>=0\forall m\)
=>Phương trình luôn có hai nghiệm
Để phương trình có hai nghiệm phân biệt thì \(\Delta>0\)
=>\(\left(m+3\right)^2>0\)
=>\(m+3\ne0\)
=>\(m\ne-3\)
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\left(m+2\right)\\x_1x_2=\dfrac{c}{a}=-2m-5\end{matrix}\right.\)
\(\left|x_1\right|+\left|x_2\right|=2\)
=>\(x_1^2+x_2^2+2\left|x_1x_2\right|=4\)
=>\(\left(x_1+x_2\right)^2-2x_1x_2+2\left|x_1x_2\right|-4=0\)
=>\(\left(2m+4\right)^2-2\left(-2m-5\right)+2\left|-2m-5\right|-4=0\)
=>\(4m^2+16m+16+4m+10+2\left|2m+5\right|-4=0\)
=>\(4m^2+20m+22+2\left|2m+5\right|=0\)(1)
TH1: m>=-5/2
Phương trình(1) sẽ trở thành
\(4m^2+20m+22+4m+10=0\)
=>\(4m^2+24m+32=0\)
=>\(m^2+6m+8=0\)
=>(m+2)(m+4)=0
=>\(\left[{}\begin{matrix}m=-2\\m=-4\end{matrix}\right.\)
TH2: m<-5/2
Phương trình (1) sẽ trở thành:
\(4m^2+20m+22-4m-10=0\)
=>\(4m^2+16m+12=0\)
=>(m+1)(m+3)=0
=>\(\left[{}\begin{matrix}m=-1\left(nhận\right)\\m=-3\left(loại\right)\end{matrix}\right.\)
\(x^2-2\left(m+2\right)-2m-5=0\)
\(\Delta=\left[-2\left(m+2\right)\right]^2-4\cdot1\cdot\left(-2m-5\right)=4\left(m^2+4m+4\right)+4\left(2m+5\right)\)
\(=4m^2+24m+36=\left(2m+6\right)^2\ge0\forall m\)
Để pt có 2 nghiệm phân biệt thì \(2m+6\ne0\Leftrightarrow m\ne-3\)
Theo vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+2\right)\\x_1x_2=-2m-5\end{matrix}\right.\)
\(\left|x_1\right|+\left|x_2\right|=2\Leftrightarrow\sqrt{\left(\left|x_1\right|+\left|x_2\right|\right)^2}=2\)
\(\Leftrightarrow\sqrt{x_1^2+x_2^2+2\left|x_1x_2\right|}=2\)
\(\Leftrightarrow\sqrt{\left(x_1+x_2\right)^2-2x_1x_2+2\left|x_1x_2\right|}=2\)
\(\Leftrightarrow\sqrt{4\left(m+2\right)^2-2\cdot\left(-2m-5\right)+2\left|-2m-5\right|}=2\)
\(\Leftrightarrow4\left(m^2+4m+4\right)+4m+10+2\left|-2m-5\right|=4\)
\(\Leftrightarrow4m^2+20m+26+2\left|-2m-5\right|=4\)
\(\Leftrightarrow4m^2+20m+22+2\left|-2m-5\right|=0\)
TH1: \(x< -\dfrac{5}{2}\)
\(4m^2+20m+22-4m-10=0\Leftrightarrow4m^2+16m+12=0\Leftrightarrow\left[{}\begin{matrix}m=-1\left(ktm\right)\\m=-3\left(ktm\right)\end{matrix}\right.\)
TH2: \(x\ge-\dfrac{5}{2}\)
\(4m^2+20m+22+4m+10=0\Leftrightarrow4m^2+24m+32=0\Leftrightarrow\left[{}\begin{matrix}m=-2\left(tm\right)\\m=-4\left(ktm\right)\end{matrix}\right.\)
Vậy: ...