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

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

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