Cross Multiplication?

Let $\sqrt{x+1}$ be $a$ and $\sqrt{x-1}$ be $b$ .
Then $\dfrac{a + b}{a - b} = 3\\ a + b = 3a - 3b\\ 2a = 4b\\ a = 2b$
Squaring on both sides:-
$a^2 = 4b^2\\ x + 1 = 4x - 4\\ 3x = 5\\ x = \dfrac{5}{3}\\ x = \boxed{1.67}$ .

Relevant wiki: Componendo and Dividendo

By the 'componendo and dividendo' rule which states if : $\frac{a}{b}=\frac{c}{d}$ then $\frac{a+b}{a-b}=\frac{c+d}{c-d}.$ Therefore, : $\frac{\sqrt{x+1}+\sqrt{x-1}+\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}-\sqrt{x+1}-\sqrt{x-1}}=\frac{3+1}{3-1};$ which yields $\frac {\sqrt{x+1}}{\sqrt{x-1}}=\frac {2}{1};$ On squaring both sides we get $\frac {x+1}{x-1}=4$ now again on applying componendo and dividendo we get $\frac{{x+1}+{x-1}}{{x+1}-{x-1}}=\frac{4+1}{4-1};$ which yields $x=\frac{5}{3};$ or $x=1.67$

$\dfrac{\sqrt{x-1}+\sqrt{x+1}}{\sqrt{x+1}-\sqrt{x-1}}=\dfrac{\left(\sqrt{x-1}+\sqrt{x+1} \right )^2}{\left(\sqrt{x+1}-\sqrt{x-1}\right) \left(\sqrt{x-1}+\sqrt{x+1}\right)}$

$\cdots=\dfrac{2 \sqrt{x^2-1}+2 x}{2}=\sqrt{x^2-1}+x$

$\sqrt{x^2-1}+x=3 \Rightarrow \sqrt{x^2-1}=(3-x) \Rightarrow x^2-1=x^3-6x+9 \Rightarrow 6x=10$

$x=\boxed{\dfrac{5}{3}}$

Relevant wiki: Componendo and Dividendo

By the 'componendo and dividendo' rule which states if $\frac{a}{b}=\frac{c}{d}$ then $\frac{a+b}{a-b}=\frac{c+d}{c-d}.$
Therefore, $\frac{\sqrt{x+1}+\sqrt{x-1}+\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}-\sqrt{x+1}-\sqrt{x-1}}=\frac{3+1}{3-1};$
when simplify this we get $2\sqrt{x-1}=\sqrt{x+1}.i.e.,4(x-1)=(x+1)$ and therefore $x=\frac{5}{3}\cong \boxed{1.67}.$