ĐKXĐ: \(x\ge-\dfrac{9}{2};x\ne0\)
\(\dfrac{2x^2\left(3+\sqrt{2x+9}\right)^2}{\left(3-\sqrt{2x+9}\right)^2\left(3+\sqrt{2x+9}\right)^2}=x+9\)
\(\Leftrightarrow\dfrac{2x^2\left(3+\sqrt{2x+9}\right)^2}{4x^2}=x+9\)
\(\Leftrightarrow\left(3+\sqrt{2x+9}\right)^2=2x+18\)
Đặt \(\sqrt{2x+9}=t>0\) pt trở thành:
\(\left(3+t\right)^2=t^2+9\)
\(\Leftrightarrow6t=0\Rightarrow t=0\)
\(\Rightarrow\sqrt{2x+9}=0\Rightarrow x=-\dfrac{9}{2}\)
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