Please

Please determine the solvability conditions of the equation, solve the equation and perform the test:

x divided by x squared minus 2x plus1 the whole minus x + 3 divided by x squared minus 1 this is equal to 0:

x/(x²-2x+1) - (x+3)/( x²-1) = 0

Correct answer:

x =  3

Step-by-step explanation:

$$\begin{gathered} {\displaystyle \frac{x}{x^2-2x+1}}-{\displaystyle \frac{x+3}{x^2-1}}=0 \\ {k}^2-2k+1=0 \\ {k}^2-2k+1=0 \\ {k}^2-2k+1=0 \\ a=1;b=-2;c=1 \\ D=b^2-4ac=2^2-4\cdot 1\cdot 1=0 \\ D=0 \\ {k}_{1,2}={\displaystyle \frac{-b\pm \sqrt{D}}{2a}}={\displaystyle \frac{2\pm \sqrt{0}}{2}}=1\pm \sqrt{0} \\ {k}_{1,2}=1\pm 0 \\ {k}_1=k_2=1 \\ \text{ Factored form of the equation: } \\ (k-1)(k-1)=0 \\ {q}^2-1=0 \\ {q}^2-1=0 \\ {q}^2-1=0 \\ a=1;b=0;c=-1 \\ D=b^2-4ac=0^2-4\cdot 1\cdot (-1)=4 \\ D>0 \\ {q}_{1,2}={\displaystyle \frac{-b\pm \sqrt{D}}{2a}}={\displaystyle \frac{\pm \sqrt{4}}{2}} \\ {q}_{1,2}={\displaystyle \frac{\pm 2}{2}} \\ {q}_{1,2}=\pm 1 \\ {q}_1=1 \\ {q}_2=-1 \\ \text{ Factored form of the equation: } \\ (q-1)(q+1)=0 \\ x\mathrm{\ne }{k}_1,k_1,q_1,q_2 \\ x\mathrm{\ne }+1;x\mathrm{\ne }-1 \\ {\displaystyle \frac{x}{x^2-2x+1}}-{\displaystyle \frac{x+3}{x^2-1}}=0 \\ {\displaystyle \frac{x}{(x-1)(x-1)}}-{\displaystyle \frac{x+3}{(x+1)(x-1)}}=0 \\ {\displaystyle \frac{x}{(x-1)}}-{\displaystyle \frac{x+3}{(x+1)}}=0 \\ x\cdot (x+1)-(x+3)\cdot (x-1)=0 \\ x=3 \\ \text{ Verifying Solution: } \\ T={\displaystyle \frac{x}{x^2-2x+1}}-{\displaystyle \frac{x+3}{x^2-1}}={\displaystyle \frac{3}{3^2-2\cdot 3+1}}-{\displaystyle \frac{3+3}{3^2-1}}=0 \end{gathered}$$

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