g) \( \frac{a-2}{a^{2}-1}-\frac{a+1}{a^{2}-2 a+1} \)

To simplify the expression \( \frac{a-2}{a^{2}-1}-\frac{a+1}{a^{2}-2a+1} \), we start by factoring the denominators. The first denominator, \( a^{2}-1 \), can be factored using the difference of squares: \[ a^2 - 1 = (a-1)(a+1) \] The second denominator, \( a^{2}-2a+1 \), is a perfect square: \[ a^{2}-2a+1 = (a-1)^2 \] Now, we rewrite the original expression with the factored denominators: \[ \frac{a-2}{(a-1)(a+1)} - \frac{a+1}{(a-1)^2} \] Next, we will find a common denominator to combine the fractions. The common denominator is \( (a-1)^2(a+1) \). Now, we rewrite each fraction with the common denominator: 1. The first fraction: \[ \frac{(a-2)(a-1)}{(a-1)(a+1)(a-1)} = \frac{(a-2)(a-1)}{(a-1)^2(a+1)} \] 2. The second fraction: \[ \frac{(a+1)(a+1)}{(a-1)^2(a+1)} = \frac{(a+1)^2}{(a-1)^2(a+1)} \] Now we put these together: \[ \frac{(a-2)(a-1) - (a+1)^2}{(a-1)^2(a+1)} \] Now, let's simplify the numerator: \[ (a-2)(a-1) - (a^2 + 2a + 1) \] Expanding \( (a-2)(a-1) \): \[ = a^2 - a - 2a + 2 = a^2 - 3a + 2 \] Now we combine that with the second part: \[ a^2 - 3a + 2 - (a^2 + 2a + 1) = a^2 - 3a + 2 - a^2 - 2a - 1 \] \[ = -5a + 1 \] Now, the overall expression simplifies to: \[ \frac{-5a + 1}{(a-1)^2(a+1)} \] Thus, the final simplified expression is: \[ \frac{1 - 5a}{(a-1)^2(a+1)} \]