Question

Giúp mình với ạ

Answer 1

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)

$\Leftrightarrow x=\frac{77\pm 9\sqrt{73}}{8}$
 

Answer 2

a: Ta có: \(2x-9\sqrt{x}+1=0\)

\(\Leftrightarrow x-2\cdot\sqrt{x}\cdot\dfrac{9}{4}+\dfrac{81}{16}-\dfrac{65}{16}=0\)

\(\Leftrightarrow\left(\sqrt{x}-\dfrac{9}{4}\right)^2=\dfrac{65}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{\sqrt{65}+9}{4}\\\sqrt{x}=\dfrac{-\sqrt{65}+9}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{73+9\sqrt{65}}{8}\\x=\dfrac{73-9\sqrt{65}}{8}\end{matrix}\right.\)