smc

First, square both sides. This gives us $${\sqrt{5x-1}}^2+2\cdot \sqrt{5x-1}\cdot \sqrt{x-1}+{\sqrt{x-1}}^2=4⟹5x-1+2\cdot \sqrt{(5x-1)\cdot (x-1)}+x-1=4⟹2\cdot \sqrt{5x^2-6x+1}+6x-2=4$$ Then, adding $-6x$ to both sides gives us $$2\cdot \sqrt{5x^2-6x+1}+6x-2-6x=4-6x⟹2\cdot \sqrt{5x^2-6x+1}-2=-6x+4$$ After that, adding $2$ to both sides will give us $$2\cdot \sqrt{5x^2-6x+1}-2+2=-6x+4+2⟹2\cdot \sqrt{5x^2-6x+1}=-6x+6$$ Next, we divide both sides by 2 which gives us $$\frac{2\cdot \sqrt{5x^2-6x+1}-2+2}{2}=\frac{-6x+4+2}{2}⟹\sqrt{5x^2-6x+1}=-3x+3$$ Finally, solving the equation, we get $$5x^2-6x+1=(-3x+3{)}^2⟹5x^2-6x+1=9x^2-18x+9$$ $$⟹5x^2-6x+1-(9x^2-18x+9)=9x^2-18x+9-(9x^2-18x+9)$$ $$⟹-4x^2+12x-8=0⟹-4(x-1)(x-2)=0$$ $$⟹x-1=0\text{or}x-2=0⟹x=1\text{or}x=2$$ Plugging 1 and 2 into the original equation, $\sqrt{5x-1}+\sqrt{x-1}=2$, we see that when $x=1$ $$\sqrt{5x-1}+\sqrt{x-1}=2⟹\sqrt{5\cdot 1-1}+\sqrt{1-1}=2⟹\sqrt{4}+\sqrt{0}=2⟹2+0=2⟹2=2$$ the equation is true. On the other hand, we note that when $x=2$ $$\sqrt{5x-1}+\sqrt{x-1}=2⟹\sqrt{5\cdot 2-1}+\sqrt{2-1}=2⟹\sqrt{9}+\sqrt{1}=2⟹3+1=2⟹4=0$$ the equation is false. Therefore the answer is $\boxed{x=1}$.