$x+\frac{x}{\sqrt{x^{2}-1}}=\frac{35}{12}$
C1:Lượng giác hóa 
Đk:$\begin{cases}x^2-1>0 \\ x>0 \end{cases}\Leftrightarrow x>1$
Đặt $x=\frac{1}{sint}(*),t\in (0;\frac{\pi}{2})\Rightarrow \begin{cases}sint>0 \\ cost>0 \end{cases}$
Pt $\Leftrightarrow sint+cost=\frac{35}{12}sintcost  (1)$
Đặt $a=sint+cost,a\in (1;\sqrt{2}]\Rightarrow \frac{a^2-1}{2}=sintcost$
$(1)\Leftrightarrow 35a^2-24a-35=0\Leftrightarrow a=\frac{7}{5}\vee a=-\frac{5}{7}(loại)$
$\Rightarrow t=sin^{-1}\frac{7}{5\sqrt{2}}-\frac{\pi}{4}\vee t=\frac{3\pi}{4}-sin^{-1}\frac{7}{5\sqrt{2}}$
Thế vào $(Shift8)$ ta được $x=\frac{5}{3}\vee x=\frac{5}{4}$
C2: Đặt ẩn phụ
Đặt $\begin{cases}a=x>1 \\ b=\sqrt{x^2-1}>0 \end{cases}\Rightarrow \begin{cases}a^2-b^2=1 \\ a(b+1)=\frac{35}{12}b \end{cases}$
$\Leftrightarrow \begin{cases}a=\frac{35b}{12(b+1)} \\ \frac{1225b^2}{144(b^2+2b+1)}-b^2=1 \end{cases}\Leftrightarrow \begin{cases}a=\frac{35b}{12(b+1)},a>1,b>0 \\ 144b^4+288b^3-937b^2+288b+144=0 \end{cases}$
$\Leftrightarrow (b-\frac{4}{3})(b-\frac{3}{4})(b+\frac{49-5\sqrt{73}}{24})(b+\frac{49+5\sqrt{73}}{24})=0$
$\Leftrightarrow b=\frac{4}{3}\vee b=\frac{3}{4}$
Thế vào ta được $x=\frac{5}{3}\vee x=\frac{5}{4}$


công nhận bạn nhậm nghiệm giỏi thật bày mình với –  danhvietuong1998 12-06-14 04:29 PM
Thấy đúng thì nhấn vào biểu tượng chữ V màu trắng bên dưới vote down nhé! Lần sau mình sẽ ss giúp đỡ. Tks bạn –  Nero 11-06-14 09:23 PM
Điều kiện: $x>1$
Phương trình tương đương với:
     $(x+\dfrac{x}{\sqrt{x^2-1}})^2=\dfrac{1225}{144}$
$\Leftrightarrow x^2+\dfrac{2x^2}{\sqrt{x^2-1}}+\dfrac{x^2}{x^2-1}-\dfrac{1225}{144}=0$
$\Leftrightarrow \dfrac{x^4}{x^2-1}+\dfrac{2x^2}{\sqrt{x^2-1}}-\dfrac{1225}{144}=0$
$\Leftrightarrow \dfrac{x^2}{\sqrt{x^2-1}}=\dfrac{25}{12}$
$\Leftrightarrow 12x^2=25\sqrt{x^2-1}$
$\Leftrightarrow 144x^4-625x^2+625=0$
$\Leftrightarrow \left[ \begin{array}{l} x=\dfrac{5}{4}\\x=\dfrac{5}{3} \end{array} \right.$

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