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 one is equal to 0:

x/(x²-2x+1) - (x+3)/( x²-1) = 0

Final Answer:

x =  3

Step-by-step explanation:

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

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