\(xét^{ }x^2=mx+5
=>x^2-mx-5=0\)
đen ta >0
<=> \(m^2.\left(-4\right).\left(-5\right)=m^2+20>0\)
vì x1<x2
=> x2= \(\frac{m+\sqrt{m^2+20}}{2}\)
=> I x2I = \(\orbr{\orbr{\begin{cases}x=\frac{m+\sqrt{m^2+20}}{2}\\x=\frac{-m-\sqrt{m^2+20}}{2}\end{cases}}}\)( với m< \(\sqrt{m^2+20}\))và (m >\(\sqrt{m^2+20}\)
=> x1=\(\frac{m-\sqrt{m^2+20}}{2}\)
=> Ix1I =\(\orbr{\begin{cases}\frac{m-\sqrt{m^2+20}}{2}\\\frac{-m+\sqrt{m^2+20}}{2}\end{cases}}\)
vì Ix1I >Ix2I <=> Ix1I -Ix2I >0
trường hợp 1:
x1-x2= \(\frac{m-\sqrt{m^2+20}}{2}\)-\(\frac{m+\sqrt{m^2+20}}{2}\)= \(\frac{m-\sqrt{m^2+20}}{2}\frac{-m-\sqrt{m^2+20}}{2}=\frac{-2\sqrt{m+20}}{2}=-\sqrt{m^2+20}>0\)(vô lí)
trường hợp 2
x1-x2= \(\frac{m-\sqrt{m^2+20}}{2}\)- \(\frac{-m-\sqrt{m^2+20}}{2}\)= \(\frac{m-\sqrt{m^2+20}}{2}\frac{+m+\sqrt{m^2+20}}{2}=\frac{2m}{2}>0\)=> m>0
