\(\dfrac{9}{x^2}+\dfrac{2x}{\sqrt{2x^2+9}}=1\left(x\ne0\right)\)
Đặt: \(\sqrt{2x^2+9}=a\left(a\ge0\right)\)
\(\Rightarrow2x^2+9=a^2\Leftrightarrow9=a^2-2x^2\)
Khi đó, phương trình đã cho trở thành:
\(\dfrac{a^2-2x^2}{x^2}+\dfrac{2x}{a}=1\)
\(\Leftrightarrow\left(\dfrac{a}{x}\right)^2-2+\dfrac{2x}{a}-1=0\)
\(\Leftrightarrow\left(\dfrac{a}{x}\right)^2+\dfrac{2x}{a}-3=0\left(1\right)\)
Đặt: \(\dfrac{a}{x}=b\), khi đó (1) trở thành:
\(b^2+\dfrac{2}{b}-3=0\)
\(\Leftrightarrow b^3+2-3b=0\)
\(\Leftrightarrow\left(b^3-b\right)-\left(2b-2\right)=0\)
\(\Leftrightarrow b\left(b+1\right)\left(b-1\right)-2\left(b-1\right)=0\)
\(\Leftrightarrow\left(b-1\right)\left(b^2+b-2\right)=0\)
\(\Leftrightarrow\left(b-1\right)^2\left(b+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}b-1=0\\b+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}b=1\\b=-2\end{matrix}\right.\)
Với \(b=1\), ta có:
\(\dfrac{a}{x}=1\Leftrightarrow a=x\Leftrightarrow\sqrt{2x^2+9}=x\Leftrightarrow2x^2+9=x^2\)
\(\Leftrightarrow x^2+9=0\left(loại\right)\)
Với \(b=-2\), ta có:
\(\dfrac{a}{x}=-2\Leftrightarrow a=-2x\Leftrightarrow\sqrt{2x^2+9}=-2x\)
\(\Leftrightarrow2x^2+9=4x^2\)
\(\Leftrightarrow2x^2=9\)
\(\Leftrightarrow x^2=\dfrac{9}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{\sqrt{2}}\\x=-\dfrac{3}{\sqrt{2}}\end{matrix}\right.\)
Thử lại thấy: \(x=\dfrac{3}{\sqrt{2}}\left(ktm\right);x=-\dfrac{3}{\sqrt{2}}\left(tm\right)\)
Vậy nghiệm của phương trình là \(S=\left\{-\dfrac{3}{\sqrt{2}}\right\}\)
