General Principles of Solving Radical Equations

One rather general strategy is to replace each new root $\sqrt[k]{expression}$ in the equation by a new variable, $r_j$, together with a new equation $r_j^k = expression$ (so now you will have $m+1$ polynomial equations in $m+1$ unknowns, where $m$ is the number of roots). Then eliminate variables from the system, ending with a single polynomial equation in one unknown, such that your original variable can be expressed in terms of the roots of this polynomial. This procedure can introduce spurious solutions if you only want the principal branch of the $k$'th root, so don't forget to check whether the solutions you get are valid.

For example, in your second equation, we get the system $$ \eqalign{r_1 + r_2 + r_3 - 2 \sqrt{2} &= 0\cr r_1^2-(3x-1) &= 0\cr r_2^2-(5x-3) &= 0\cr r_3^2-(x-1) &=0\cr}$$ Take the resultant of the first two polynomials with respect to $r_1$, then the resultant of this and the third with respect to $r_2$, and the resultant of this and the fourth with respect to $r_3$. We get $$ 121 x^4-4820 x^3+28646 x^2-45364 x+21417$$ which happens to factor as $$ \left( x-1 \right) \left( x-33 \right) \left( 121\,{x}^{2}-706\,x+ 649 \right) $$ However, only the solution $x=1$ turns out to satisfy the original equation.

I would write the equation in the form $$\sqrt{x+1}+\sqrt{x-1}=x+\sqrt{x^2-1}$$ with $$x\geq 1$$ after squaring one times and isolating the square root we get $$2\sqrt{1-x^2}(1-x)=2x^2-2x-1$$ squaring again we obtain $$(x^2-1)(2-2x)^2=(2x^2-2x-1)^2$$ this gives the equation $$4x-5=0$$ and we get $$x=\frac{5}{4}$$ fulfills our equation. in the second equation we get $$x=1$$ write your third equation in the form $$\sqrt{\frac{4x+1}{x+3}}=1+\sqrt{\frac{x-2}{x+3}}$$ we get after squaring two times $$\left(\frac{2x}{x+3}\right)^2-4\left(\frac{x-2}{x+3}\right)=0$$ after simplifying we obtain$$-4\,{\frac {x-6}{ \left( x+3 \right) ^{2}}}=0$$