Lời giải:
ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow 2(x-4,5\sqrt{x}+2,25^2)-\frac{73}{8}=0$
$\Leftrightarrow 2(\sqrt{x}-2,25)^2=\frac{73}{8}$
$\Leftrightarrow (\sqrt{x}-2,25)^2=\frac{73}{16}$
$\Rightarrow \sqrt{x}-2,25=\frac{\pm \sqrt{73}}{4}$
$\Leftrightarrow \sqrt{x}=\frac{9\pm \sqrt{73}}{4}$ (đều tm)
$\Leftrightarrow x=\frac{77\pm 9\sqrt{73}}{8}$
a: Ta có: \(2x-9\sqrt{x}+1=0\)
\(\Leftrightarrow x-2\cdot\sqrt{x}\cdot\dfrac{9}{4}+\dfrac{81}{16}-\dfrac{65}{16}=0\)
\(\Leftrightarrow\left(\sqrt{x}-\dfrac{9}{4}\right)^2=\dfrac{65}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{\sqrt{65}+9}{4}\\\sqrt{x}=\dfrac{-\sqrt{65}+9}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{73+9\sqrt{65}}{8}\\x=\dfrac{73-9\sqrt{65}}{8}\end{matrix}\right.\)