giải phương trình [tex]\sqrt{x^2+1} + \sqrt{2x^2+4x+3}=2x+1[/tex]
[tex]\sqrt{x^2+1} + \sqrt{2x^2+4x+3}=2x+1\\\Leftrightarrow (\sqrt{x^2+1}-x)+[\sqrt{2x^2+4x+3}-(x+1)]=0\\\Leftrightarrow \frac{(x^2+1)-x^2}{\sqrt{x^2+1}+x}+\frac{2x^2+4x+3-(x+1)^2}{\sqrt{2x^2+4x+3}+(x+1)}=0\\\Leftrightarrow \frac{1}{\sqrt{x^2+1}+x}+\frac{x^2+2x+2}{\sqrt{2x^2+4x+3}+(x+1)}=0[/tex]
Vì [tex]\sqrt{x^2+1} + \sqrt{2x^2+4x+3}>0[/tex] mà [TEX]\sqrt{x^2+1} + \sqrt{2x^2+4x+3}=2x+1[/TEX]
Nên [tex]2x+1>0\Leftrightarrow x>\frac{-1}{2}[/tex]
Ta có
- [tex]\sqrt{x^2+1}+x>\sqrt{\left ( \frac{-1}{2} \right )^2+1}-\frac{1}{2}>0\Rightarrow \frac{1}{\sqrt{x^2+1}+x}>0[/tex]
- [tex]\left\{\begin{matrix} x^2+2x+2>0\\\sqrt{2x^2+4x+3}+(x+1)>0+\frac{-1}{2}+1>0(vì:2x^2+4x+3=2(x+1)^2+1>0) \end{matrix}\right.\\\Rightarrow \frac{x^2+2x+2}{\sqrt{2x^2+4x+3}+(x+1)}>0[/tex]
Suy ra [tex]\frac{1}{\sqrt{x^2+1}+x}+\frac{x^2+2x+2}{\sqrt{2x^2+4x+3}+(x+1)}>0[/tex]
Vậy pt đã cho vô nghiệm.
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
