ĐK: $x> 1$
$\frac{-1}{x^2}-\frac{-1}{x}+\frac{1}{x-1}=\frac{4(\sqrt{4x-3}+1)}{x+\sqrt{x^2+x}}$
$\Leftrightarrow \frac{1}{x-1}-\frac{1}{x}-4=\frac{1}{x^2}-\frac{4x-4}{x}+\frac{4(\sqrt{4x-3}+1)}{x+\sqrt{x^2+x}}-\frac{4}{x}$
$\frac{1-4x^2+4x}{x(x-1)}=\frac{1-4x^2+4x}{x^2}+4.\frac{x\sqrt{4x-3}+x-x-\sqrt{x^2+x}}{x(x+\sqrt{x^2+x})}$
$(4x^2+4x-1)[\frac{4}{(x+\sqrt{x^2+x})(\sqrt{4x-3}+\sqrt{x^2+x})}-\frac{1}{x^2}+\frac{1}{x(x-1)}]=0$
phần trong ngoặc luôn (+) vì
$\frac{-1}{x^2}+\frac{1}{x(x-1)}=\frac{1}{x^2(x-1)}>0 $ với $x> 1$
vậy nghiệm pt là $\frac{1+\sqrt{2}}{2}$
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