1. **State the problem:** Simplify and solve the equation $$x^2 - x - \frac{x^2 - x + 2}{x^2 + 1} + \frac{x^2}{x - 2} = 1$$ for $x$. 2. **Rewrite the equation:** To solve, get all terms on one side: $$x^2 - x - \frac{x^2 - x + 2}{x^2 + 1} + \frac{x^2}{x - 2} - 1 = 0$$ 3. **Combine like terms:** Note that $x^2 - x - 1$ can be grouped: $$\left(x^2 - x - 1\right) - \frac{x^2 - x + 2}{x^2 + 1} + \frac{x^2}{x - 2} = 0$$ 4. **Find a common denominator:** The denominators are $1$, $x^2 + 1$, and $x - 2$. Let's find a common denominator of $(x^2 + 1)(x - 2)$. Multiply each term accordingly: $$\frac{(x^2 - x - 1)(x^2 + 1)(x - 2)}{(x^2 + 1)(x - 2)} - \frac{(x^2 - x + 2)(x - 2)}{(x^2 + 1)(x - 2)} + \frac{x^2(x^2 + 1)}{(x^2 + 1)(x - 2)} = 0$$ 5. **Combine over the common denominator:** $$\frac{(x^2 - x - 1)(x^2 + 1)(x - 2) - (x^2 - x + 2)(x - 2) + x^2(x^2 + 1)}{(x^2 + 1)(x - 2)} = 0$$ 6. **Set numerator equal to zero:** Since the denominator cannot be zero for solutions, set numerator to zero to find the roots: $$ (x^2 - x - 1)(x^2 + 1)(x - 2) - (x^2 - x + 2)(x - 2) + x^2(x^2 + 1) = 0 $$ 7. **Expand each part:** - Expand $(x^2 - x - 1)(x^2 + 1)$: $$ (x^2 - x - 1)(x^2 + 1) = x^4 - x^3 - x^2 + x^2 - x - 1 = x^4 - x^3 - x - 1 $$ - Multiply by $(x - 2)$: $$ (x^4 - x^3 - x - 1)(x - 2) = x^5 - 2x^4 - x^4 + 2x^3 - x^2 + 2x - x + 2 = x^5 - 3x^4 + 2x^3 - x^2 + x + 2 $$ - Expand $-(x^2 - x + 2)(x - 2)$: $$ -(x^3 - 2x^2 - x^2 + 2x + 2x -4) = -(x^3 - 3x^2 + 4x -4) = -x^3 + 3x^2 - 4x + 4 $$ - Expand $x^2(x^2 + 1) = x^4 + x^2$ 8. **Sum all expanded terms:** $$ x^5 - 3x^4 + 2x^3 - x^2 + x + 2 - x^3 + 3x^2 - 4x + 4 + x^4 + x^2 = 0 $$ 9. **Combine like terms:** $$ x^5 + (-3x^4 + x^4) + (2x^3 - x^3) + (-x^2 + 3x^2 + x^2) + (x - 4x) + (2 + 4) = 0 $$ $$ x^5 - 2x^4 + x^3 + 3x^2 - 3x + 6 = 0 $$ 10. **Solve the quintic polynomial:** The equation to solve is $$ x^5 - 2x^4 + x^3 + 3x^2 - 3x + 6 = 0 $$ This is a nontrivial polynomial equation of degree 5, unsolvable by radicals generally. We check for rational roots with Rational Root Theorem: possible roots $\\pm 1, \\pm 2, \\pm 3, \\pm 6$. Evaluate at $x=1$: $1 - 2 + 1 + 3 - 3 + 6 = 6 \neq 0$ At $x = -1$: $-1 - 2(-1)^4 + (-1)^3 + 3(-1)^2 - 3(-1) + 6 = -1 - 2 -1 + 3 + 3 + 6 = 8 \neq 0$ At $x=2$: $32 - 2(16) + 8 + 3(4) - 6 + 6 = 32 - 32 + 8 + 12 - 6 + 6 = 20 \neq 0$ At $x=-2$: $-32 - 2(16) - 8 + 3(4) + 6 + 6 = -32 - 32 - 8 + 12 + 6 + 6 = -48 \neq 0$ No obvious rational roots; numeric or graphing methods required for approximate solutions. **Final answer:** The original equation reduces to a fifth-degree polynomial: $$ x^5 - 2x^4 + x^3 + 3x^2 - 3x + 6 = 0 $$ Solutions to this equation can be approximated only numerically. ---