$$\sqrt{x^2-2x+5} + 2\sqrt{4x+5} = x^3-2x^2+5x+4$$
$$\iff (\sqrt{x^2-2x+5}-2) + 2(\sqrt{4x+5}-3) = x^3-2x^2+5x-4$$
$$\iff \dfrac{(x-1)^2}{\sqrt{x^2-2x+5}+2}+\dfrac{8(x-1)}{\sqrt{4x+5}+3}=(x-1)(x^2-x+4)$$
$$\iff \begin{bmatrix}
x=1 & \\
\dfrac{x-1}{\sqrt{x^2-2x+5}+2}+\dfrac{8}{\sqrt{4x+5}+3}=x^2-x+4&
\end{bmatrix}$$
Xét: $$\dfrac{x-1}{\sqrt{x^2-2x+5}+2}+\dfrac{8}{\sqrt{4x+5}+3}=x^2-x+4$$
• $x > 1$
$\sqrt{x^2-2x+5}+2 =\sqrt{(x-1)^2+4}+2 > 4 \Longrightarrow \dfrac{x-1}{\sqrt{x^2-2x+5}+2} < \dfrac{x-1}{4}$
$\sqrt{4x+5}+3=\sqrt{4(x-1)+9}+3 > 6 \Longrightarrow \dfrac{8}{\sqrt{4x+5}+3} < \dfrac{4}{3}$
$$ \Longrightarrow x^2-x+4=\dfrac{x-1}{\sqrt{x^2-2x+5}+2}+\dfrac{8}{\sqrt{4x+5}+3} < \dfrac{x-1}{4}+ \dfrac{4}{3}$$
$$ \iff x^2-\dfrac{5x}{4}+\dfrac{35}{12}<0$$ (vô lý)
• $\dfrac{-5}{4} \le x<1$
$ \dfrac{x-1}{\sqrt{x^2-2x+5}+2} < 0$
$\sqrt{4x+5}+3 >3 \Longrightarrow \dfrac{8}{\sqrt{4x+5}+3} < \dfrac{8}{3}$
$VT < \dfrac{8}{3}$
$VP=x^2-x+4=(x-\dfrac{1}{2})^2+\dfrac{15}{4} > \dfrac{8}{3}$
vậy pt có nghiệm duy nhất $x=1$
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