coi kỹ nha sợ nhầm khúc nào á:v
Ta có :
ac = -2015 . 2013 < 0 nên pt đã cho luôn có 2 nghiệm pb x1 ; x2
có:
\(\sqrt{x^2_1+2014}-x_1=\sqrt{x^2_2+2014}+x_2\)
<=>
\(\sqrt{x^2_2+2014}-\sqrt{x^2_1+2014}+x_1+x_1=0\)
<=>
\(\dfrac{x^2_2-x^2_1}{\sqrt{x^2_2+2014}+\sqrt{x^2_1+2014}}+x_2+x_1=0\)
<=>
\(\left(x_2+x_1\right)\left(\dfrac{x_2-x_1}{\sqrt{x^2_2+2014}+\sqrt{x^2_1+2014}}+1\right)=0\)
<=>
\(\left[{}\begin{matrix}x_2+x_1=0\\x_2-x_1+\sqrt{x^2_2+2014}+\sqrt{x^2_1+2014}=0\left(1\right)\end{matrix}\right.\)
Ta có : x2 + x1 = 0
<=>
\(\dfrac{m-2014}{2013}=0\Leftrightarrow m=2014\)
\(\sqrt{x^2_2+2014}+\sqrt{x^2_1+2014}>\sqrt{x^2_2}+\sqrt{x^2_1}=\left|x_2\right|+\left|x_1\right|\ge-x_2+x_1\)
suy ra:
\(\sqrt{x^2_2+2014}+\sqrt{x^2_1+2014}+x_2-x_1>0\)
=> (1) vô nghiệm
Vậy giá trị của m là 2014
\(2013x^2-\left(m-2014\right)x-2015=0\)
\(ac=-2015.2-13< 0\)\(\Rightarrow pt\) \(luôn\) \(có\) \(2ngo\) \(pb\) \(trái\) \(dấu\)
\(\sqrt{x1^2+2014}-x1=\sqrt{x2^2+2014}+x2\)
\(\Leftrightarrow\sqrt{x1^2+2014}-x2=\sqrt{x2^2+2014}+x1\)
\(\Leftrightarrow x1^2+x2^2+2014-2x2\sqrt{x1^2+2014}=x1^2+x2^2+2014+2x1\sqrt{x2^2+2014}\)
\(\Leftrightarrow-2x2\sqrt{x1^2+2014}=2x1\sqrt{x2^2+2014}\)
\(\Leftrightarrow-x2\sqrt{x1^2+2014}=x1\sqrt{x2^2+2014}\left(1\right)\)
\(do\) \(pt\) \(có\) \(2ngo\) \(trái\) \(dấu\) \(không\) \(mất\) \(tổng\) \(quát\)
\(giả\) \(sử:x2< 0< x1\)
\(\Rightarrow\left(1\right)\Leftrightarrow x2^2\left(x1^2+2014\right)=x1^2\left(x2^2+2014\right)\Leftrightarrow x2^2=x1^2\Leftrightarrow x2=-x1\)
\(\Rightarrow\left\{{}\begin{matrix}x2=-x1\\x1+x2=\dfrac{m-2014}{2013}\end{matrix}\right.\)\(\Leftrightarrow\dfrac{m-2014}{2013}=0\Leftrightarrow m=2014\)
