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

Ask by Fuentes Crawford. in Nepal
Jan 23,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The simplified form of the expression is \( \frac{-5a+1}{a^{3}-a^{2}-a+1} \).

Solution

Calculate or simplify the expression \( (a-2)/(a^2-1)-(a+1)/(a^2-2*a+1) \). Simplify the expression by following steps: - step0: Solution: \(\frac{\left(a-2\right)}{\left(a^{2}-1\right)}-\frac{\left(a+1\right)}{\left(a^{2}-2a+1\right)}\) - step1: Remove the parentheses: \(\frac{a-2}{a^{2}-1}-\frac{a+1}{a^{2}-2a+1}\) - step2: Factor the expression: \(\frac{a-2}{\left(a+1\right)\left(a-1\right)}-\frac{a+1}{\left(a-1\right)\left(a-1\right)}\) - step3: Reduce fractions to a common denominator: \(\frac{\left(a-2\right)\left(a-1\right)}{\left(a+1\right)\left(a-1\right)\left(a-1\right)}-\frac{\left(a+1\right)\left(a+1\right)}{\left(a-1\right)\left(a-1\right)\left(a+1\right)}\) - step4: Multiply: \(\frac{\left(a-2\right)\left(a-1\right)}{\left(a+1\right)\left(a-1\right)^{2}}-\frac{\left(a+1\right)\left(a+1\right)}{\left(a-1\right)\left(a-1\right)\left(a+1\right)}\) - step5: Multiply: \(\frac{\left(a-2\right)\left(a-1\right)}{\left(a+1\right)\left(a-1\right)^{2}}-\frac{\left(a+1\right)\left(a+1\right)}{\left(a-1\right)^{2}\left(a+1\right)}\) - step6: Rewrite the expression: \(\frac{\left(a-2\right)\left(a-1\right)}{\left(a+1\right)\left(a-1\right)^{2}}-\frac{\left(a+1\right)\left(a+1\right)}{\left(a+1\right)\left(a-1\right)^{2}}\) - step7: Transform the expression: \(\frac{\left(a-2\right)\left(a-1\right)-\left(a+1\right)\left(a+1\right)}{\left(a+1\right)\left(a-1\right)^{2}}\) - step8: Multiply the terms: \(\frac{a^{2}-3a+2-\left(a+1\right)\left(a+1\right)}{\left(a+1\right)\left(a-1\right)^{2}}\) - step9: Multiply the terms: \(\frac{a^{2}-3a+2-\left(a^{2}+2a+1\right)}{\left(a+1\right)\left(a-1\right)^{2}}\) - step10: Calculate: \(\frac{-5a+1}{\left(a+1\right)\left(a-1\right)^{2}}\) - step11: Expand the expression: \(\frac{-5a+1}{a^{3}-a^{2}-a+1}\) The simplified form of the expression \( \frac{a-2}{a^{2}-1}-\frac{a+1}{a^{2}-2a+1} \) is \( \frac{-5a+1}{a^{3}-a^{2}-a+1} \).

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Bonus Knowledge

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: \[ \frac{1 - 5a}{(a-1)^2(a+1)} \]

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