<s>[INST] Solve the equation $x+\frac{x}{\sqrt{x^2-1}}=\frac{35}{12}$ Solve the equation $$x+\dfrac{x}{\sqrt{x^2-1}}=\dfrac{35}{12}.$$
The equation is defined for $x\in\left(-\infty;-1\right)\cup\left(1;+\infty\right).$ Now I am thinking how to get rid of the radical in the denominator, but I can't come up with anything. Thank you!
 [/INST] Hint:
Multiply for $12\sqrt{x^2-1}$ your initial equation,
you have $$12(\sqrt{x^2-1})x+12x=35\sqrt{x^2-1} \iff 12x(\sqrt{x^2-1}+1)=35\sqrt{x^2-1}$$
and square (two times to delete the sqrt) to the left and the right.
$$144x^2(\sqrt{x^2-1}+1)^2=35^2(x^2-1)$$
$$144x^2(x^2-1+1+2\sqrt{x^2-1})=35^2(x^2-1)$$
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