Question

cho pt: x²-2(m+2)x-2m-5=0

tim m de pt co 2 nghiem pbiet x₁,x₂ thoa man |x₁| + |x₂| =2

Answer 1

\(\Delta=\left[-2\left(m+2\right)\right]^2-4\cdot1\cdot\left(-2m-5\right)\)

\(=\left(2m+4\right)^2+4\left(2m+5\right)\)

\(=4m^2+16m+16+8m+20\)

\(=4m^2+24m+36=4\left(m^2+6m+9\right)=4\left(m+3\right)^2>=0\forall m\)

=>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

=>\(\left[{}\begin{matrix}m=-1\left(nhận\right)\\m=-3\left(loại\right)\end{matrix}\right.\)

Answer 2

\(x^2-2\left(m+2\right)-2m-5=0\)

\(\Delta=\left[-2\left(m+2\right)\right]^2-4\cdot1\cdot\left(-2m-5\right)=4\left(m^2+4m+4\right)+4\left(2m+5\right)\)

\(=4m^2+24m+36=\left(2m+6\right)^2\ge0\forall m\)

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: ...