\(\sqrt{9x^2+6x+1}\) = \(\sqrt{x^2+2+\frac{1}{x}}\) ĐK: x≠0
⇔\(\sqrt{\left(3x+1\right)^2}=\sqrt{\left(x+\frac{1}{x}\right)}\)
⇔\(\left|3x+1\right|=\left|x+\frac{1}{x}\right|\)
⇔\(\left[{}\begin{matrix}3x+1=x+\frac{1}{x}\\3x+1=-x-\frac{1}{x}\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}\left[{}\begin{matrix}x=-1\\x=\frac{1}{2}\end{matrix}\right.\\4x^2+x+1=\left(2x+\frac{1}{4}\right)^2+\frac{15}{16}>0\left(KTM\right)\end{matrix}\right.\) Vậy phương trình có nghiệm là x∈ \(\left\{-1;\frac{1}{2}\right\}\).
\(\sqrt{9x^2+6x+1}=\sqrt{x+2+\frac{1}{x}}\) (x \(\ge\) -1; x \(\ne\) 0)
\(\Leftrightarrow\) \(\sqrt{\left(3x+1\right)^2}\) = \(\sqrt{\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2}\)
\(\Leftrightarrow\) 3x + 1 = \(\sqrt{x}+\frac{1}{\sqrt{x}}\)
\(\Leftrightarrow\) 3x + 1 - \(\sqrt{x}\) + \(\frac{1}{\sqrt{x}}\) = 0
\(\Rightarrow\) Pt vô nghiệm (x \(\notin\) R)
Vậy S = \(\varnothing\)
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