$ x + \frac{x}{\sqrt{x^{2} - 1}} =\frac{35}{12} $

$x(1+\frac{1}{\sqrt{x^2-1}})=\frac{35}{12}\Rightarrow x >0$ (vì $\frac{1}{\sqrt{x^2-1}}> 0,\frac{35}{12} >0)$ và $x^2-1>0$ 
Kết hợp đk $\Rightarrow x>1$
Quy đồng khử mẫu $pt \Leftrightarrow  (12x-35)\sqrt{x^2-1}+12x=0$
$\Leftrightarrow(12x-35+\frac{25}{x}).\sqrt{x^2-1}+(12x-\frac{25\sqrt{x^2-1}}{x})=0$
$\Leftrightarrow(12x^2-35x+25).\frac{\sqrt{x^2-1}}{x}+\frac{144x^4-625x^2+625}{12x^3+25x\sqrt{x^2-1}}=0$
$\Leftrightarrow (12x^2-35x+25).\frac{\sqrt{x^2-1}}{x}+(12x^2-35x+25).\frac{12x^2-35x+25}{12x^3+25x\sqrt{x^2-1}} =0$ 
$\Leftrightarrow (12x^2-35x+25)(\frac{\sqrt{x^2-1}}{x}+\frac{12x^2-35x+25}{12x^3+25x\sqrt{x^2-1}})=0$
Dễ thấy thừa số thứ hai $>0$ do $x >1$
$\Rightarrow 12x^2-35x+25=0\Rightarrow (3x-5)(4x-5)=0$
Nên phương trình có 2 nghiệm $\color{red}{x_1=\frac{5}{3},x_2=\frac{5}{4}}$
Tạm thời chưa nghĩ ra cách nào hay hơn ^^

sai 1 ket qua kia –  hathihaianh2 27-10-15 08:11 PM

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