Solve for the real value of x

Solve the equation for $x \in \mathbf{R}$

$\frac{x}{x-1}=x+ \frac{1}{x-2}$ To solve the equation, let's start by multiplying both sides by $(x-1)(x-2)$ and simplifying we get

$\frac{x}{x-1}(x-1)(x-2)$

$=(x+\frac{1}{x-2})(x-1)(x-2)$

$x\ne 1$ and $x\ne 2$

$\implies x(x-2)=x(x-1)(x-2)+x-1$

$\implies x^{2}-2x=x^{3}$

$-2x^{2}-x^{2}+2x+x-1$

$\implies x^{3}-4x^{2}+5x-1=0$

Let $f(x)=x^{3}-4x^{2}+5x-1$

Now we are dealing with a cubic equation , this can be made by replacing $x$ by $y+ \frac{4}{3}$ to get rid of $x^{2} $ term, this gives

$(y+\frac{4}{3})^{3}-4(y+\frac{4}{3})^{2}$

$+ 5(y+\frac{4}{3})-1=0$

$ y^{2}+3y^{2}(\frac{4}{3})+$

$3y(\frac{4}{3})+ (\frac{4}{3})^3$

$-4(y^{2}+2y(\frac{4}{3}))+(\frac{4}{3})^{2}$

$+5y+\frac{20}{3}-1=0$

$ y^{3}+4y^{2}+\frac{16y}{3}$

$+\frac{64}{27}-4y^{2}-\frac{32y}{3}$

$-\frac{64}{9}+5y+\frac{20}{3}-1=0$

$\implies y^{3}-\frac{y}{3}+\frac{25}{27}=0$ We get a depressed cubic equation, to solve it, we assume that $y=s-t$ as a solution to the equation

$ y^{3}+Ay+B=0$ with $A=-\frac{1}{3}$ and

$B=\frac{25}{27}$, we get

$(s-t)^{3}+ A(s-t)+ B=0$ At this point we improve the condition $A-3st=0$ following Cardano's method, thus help to remove that fourth term in the previous equality, therefore we get

$B+s^{3}-t^{3}=9$

$A-3st=0$ Writing $s=\frac{A}{3t}$ and using the first equation of the previous system, we get

$B+(\frac{A}{3t})^{3}-t^{3}=0$ To solve that , let's make a variable change using $u=t^{3}$, we get $B=\frac{A^{3}}{27u}-u=0$ Multiplying by $u$ with $u\ne 0$ gives a quadratic equation

$u^{2}-Bu-\frac{A^{3}}{27}=0$ Let's substitute $A$ and $B$ with their values we get

$u^{2}-\frac{25u}{27}+ \frac{1}{729}=0$ Solving for $u=\frac{25+(\sqrt[3]{69})}{54}$

Here we use the larger solution , the other solution also works. This gives

$t=\frac{1}{3}\sqrt[3]{\frac{25+\sqrt[3]{69}}{2}}$ and

$s=\frac{A}{3t}=-\frac{1}{3}\sqrt[3]{\frac{2}{25+\sqrt[3]{69}}}$

Hence, the solution $x=y+\frac{4}{3}=s-t+\frac{4}{3}$

$ x=\frac{4}{3}-\frac{1}{3}\sqrt[3]{\frac{2}{25+\sqrt[3]{69}}}-\frac{1}{3}\sqrt[3]{\frac{25+\sqrt[3]{69}}{2}} \approx 0.25$

Now let's be sure that it is the only solution . To make this , we calculate the derivatives

$f^{'}=3x^{2}-8x+5=(x-1)(3x-5)$

It is possible in $(\infty, 1]$ negative in $(1,\frac{5}{3})$ and possible in $\frac{5}{3}, \infty$ , since $f(t)=1$ and $f(\frac{5}{3})=\frac{23}{27}$, the function $f$ is possible in $[, \infty$ and has only one zero which is $\approx 0.24$ in $-\infty, 1$.