Olemissmath wrote: Solve the following equation in $R$ : $(x-\frac{1}{x})^\frac{1}{2}+(1-\frac{1}{x})^\frac{1}{2}=x$ We get $LHS\ge 0$ and so $x>0$ and so, in order LHS be defined, $x\ge 1$ Equation is then equivalent (squaring) to $x\ge 1$ and $2\sqrt{(x-\frac 1x)(1-\frac 1x)}=x^2-x-1+\frac 2x$ When $x\ge 1$, $RHS \ge 0$ and equation may be equivalently squared and becomes $(x^2-x-1)^2=0$ Hence the answer $\boxed{x=\frac{1+\sqrt 5}2}$

Put $a:=\sqrt{x-{1\over x}}, b:=\sqrt{1-{1\over x}}$. Then $a+b=x\land a^2-b^2=x-1$ Hence $a-b={x-1\over x}=1-{1\over x}$ Thus $2a=(a+b)+(a-b)=x+1-{1\over x}=a^2+1\iff a=1$, yielding $x^2-x-1=0$ and the answer as above.

Olemissmath wrote: Solve the following equation in $R$ : $(x-\frac{1}{x})^\frac{1}{2}+(1-\frac{1}{x})^\frac{1}{2}=x$ $x>0 $ and $1-\frac{1}{x} \ge 0\Rightarrow x\ge 1$ and $t=\frac{1}{x}\in (0,1]$, $(x-\frac{1}{x})^\frac{1}{2}+(1-\frac{1}{x})^\frac{1}{2}=x \iff t^{\frac{1}{2}}(1-t^2)^{\frac{1}{2}}+t(1-t)^{\frac{1}{2}}=1$, $1= t^{\frac{1}{2}}(1-t^2)^{\frac{1}{2}}+t(1-t)^{\frac{1}{2}}\le \frac{t+1-t^2}{2}+\frac{t^2+1-t}{2}=1$ $\Rightarrow t=1-t^2 \Rightarrow t=\frac{1-\sqrt 5}2 \Rightarrow x=\frac{1+\sqrt 5}2.$

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=77678

A very bad solution: $$\sqrt{x-\dfrac{1}{x}}+\sqrt{1-\dfrac{1}{x}}=x\implies \sqrt{\dfrac{x-1}{x}}(\sqrt{x+1}+1)=x\implies \sqrt{\dfrac{x-1}{x}}=\sqrt{x+1}-1$$At this point I had no idea what to do so I square: $\dfrac{x-1}{x}=x+2-2\sqrt{x+1}\implies 2\sqrt{x+1}=x+1+\dfrac{1}{x}$ And again: $4(x+1)=x^2+2x+3+\dfrac{2}{x}+\dfrac{1}{x^2}\implies x^4-2x^3-x^2+2x+1=0$ Obviously this has no linear factors; now for the search of the quadratic factors of this, we can use Gauss' lemma to conclude that the constant terms for both quadratics is either both $1$ or both $-1$. With some calculation we get $x^4-2x^3-x^2+2x+1=(x^2-x-1)^2=0\implies x^2-x-1=0\implies x=\dfrac{1\pm \sqrt{5}}{2}$. However $1-\frac{1}{x} \ge 0\implies x\ge 1$ so $x=\dfrac{1-\sqrt{5}}{2}$ doesn't work; this leaves $\boxed{x=\dfrac{1+\sqrt{5}}{2}}$ as the only solution.

\[\sqrt{x-\frac1x}=x-\sqrt{1-\frac1x}\implies4(x^2-x)=(x^2-x+1)^2\]$4t=(t+1)^2\implies t=1$ so $x^2-x+1=0$

Really simple, but boring solution, you will get Transform following equation in: $\sqrt{x-\frac1x}- \sqrt{1-\frac1x}=x$ and square both sides to get this ${x-\frac1x}+ 2\sqrt{({x-\frac1x})({1-\frac1x})} + 1- \frac1x = x^2 $ and after some basic transformations you will get this ${x-\frac2x + \frac{2\sqrt{x^3-x^2-x+1}}{x}}-x^2=-1$ or ${x-\frac2x + \frac{2\sqrt{x^3-x^2-x+1}}{-x}}-x^2=-1$ because $\sqrt{x^2} = x$ or $-x$ Now you have these two equations to solve (simple) and you will get $x=\dfrac{1-\sqrt{5}}{2}$ or $x=\dfrac{1+\sqrt{5}}{2}$ and now since $x>0 $ the only solution would be $x=\dfrac{1+\sqrt{5}}{2}$ and that is all