Question

1. $\frac{a^{2}}{a - b}+\frac{b^{2}}{b - a}$
2. $\frac{-3x - 9}{x^{2}+x - 6}-\frac{x + 3}{2 - x}$
3. $\frac{8x + 2}{5x - 2}+\frac{2x - 6}{2 - 5x}$
4. $\frac{2x + 14}{x^{2}+5x - 14}-\frac{2}{x - 2}$
5. $\frac{7x + 4}{x^{2}+5x + 6}-\frac{2}{x + 3}$
6. $\frac{6}{y^{2}+6y + 9}+\frac{5}{y^{2}-9}$
7. $\frac{p + 1}{p + 3}-\frac{p + 17}{p^{2}-p - 12}$
8. $\frac{3}{5 - x}-\frac{11}{x^{2}-2x - 15}$
9. $\frac{1}{x + 1}-\frac{x}{x - 2}+\frac{x^{2}+2}{x^{2}-x - 2}$

Explanation:

Step1: Simplify the first - fraction problem

For $\frac{a^{2}}{a - b}+\frac{b^{2}}{b - a}$, rewrite $\frac{b^{2}}{b - a}=-\frac{b^{2}}{a - b}$. Then $\frac{a^{2}}{a - b}-\frac{b^{2}}{a - b}=\frac{a^{2}-b^{2}}{a - b}$. Using the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$, we get $\frac{(a + b)(a - b)}{a - b}=a + b$.

Step2: Simplify the second - fraction problem

For $\frac{-3x - 9}{x^{2}+x - 6}-\frac{x + 3}{2 - x}$, first factor $x^{2}+x - 6=(x + 3)(x - 2)$. Rewrite $\frac{x + 3}{2 - x}=-\frac{x + 3}{x - 2}$. Then $\frac{-3(x + 3)}{(x + 3)(x - 2)}+\frac{x + 3}{x - 2}=\frac{-3}{x - 2}+\frac{x + 3}{x - 2}=\frac{-3+x + 3}{x - 2}=\frac{x}{x - 2}$.

Step3: Simplify the third - fraction problem

For $\frac{8x + 2}{5x - 2}+\frac{2x - 6}{2 - 5x}$, rewrite $\frac{2x - 6}{2 - 5x}=-\frac{2x - 6}{5x - 2}$. Then $\frac{8x + 2-(2x - 6)}{5x - 2}=\frac{8x + 2-2x + 6}{5x - 2}=\frac{6x+8}{5x - 2}=\frac{2(3x + 4)}{5x - 2}$.

Step4: Simplify the fourth - fraction problem

For $\frac{2x + 14}{x^{2}+5x - 14}-\frac{2}{x - 2}$, factor $x^{2}+5x - 14=(x + 7)(x - 2)$. Then $\frac{2(x + 7)}{(x + 7)(x - 2)}-\frac{2}{x - 2}=\frac{2}{x - 2}-\frac{2}{x - 2}=0$.

Step5: Simplify the fifth - fraction problem

For $\frac{7x + 4}{x^{2}+5x + 6}-\frac{2}{x + 3}$, factor $x^{2}+5x + 6=(x + 2)(x + 3)$. Then $\frac{7x + 4}{(x + 2)(x + 3)}-\frac{2(x + 2)}{(x + 2)(x + 3)}=\frac{7x + 4-2x - 4}{(x + 2)(x + 3)}=\frac{5x}{(x + 2)(x + 3)}$.

Step6: Simplify the sixth - fraction problem

For $\frac{6}{y^{2}+6y + 9}+\frac{5}{y^{2}-9}$, factor $y^{2}+6y + 9=(y + 3)^{2}$ and $y^{2}-9=(y + 3)(y - 3)$. The common denominator is $(y + 3)^{2}(y - 3)$. $\frac{6(y - 3)}{(y + 3)^{2}(y - 3)}+\frac{5(y + 3)}{(y + 3)^{2}(y - 3)}=\frac{6y-18 + 5y+15}{(y + 3)^{2}(y - 3)}=\frac{11y - 3}{(y + 3)^{2}(y - 3)}$.

Step7: Simplify the seventh - fraction problem

For $\frac{p + 1}{p + 3}-\frac{p + 17}{p^{2}-p - 12}$, factor $p^{2}-p - 12=(p - 4)(p + 3)$. Then $\frac{(p + 1)(p - 4)}{(p + 3)(p - 4)}-\frac{p + 17}{(p - 4)(p + 3)}=\frac{p^{2}-3p - 4-(p + 17)}{(p + 3)(p - 4)}=\frac{p^{2}-3p - 4-p - 17}{(p + 3)(p - 4)}=\frac{p^{2}-4p - 21}{(p + 3)(p - 4)}=\frac{(p - 7)(p+3)}{(p + 3)(p - 4)}=\frac{p - 7}{p - 4}$.

Step8: Simplify the eighth - fraction problem

For $\frac{3}{5 - x}-\frac{11}{x^{2}-2x - 15}$, factor $x^{2}-2x - 15=(x - 5)(x + 3)$. Rewrite $\frac{3}{5 - x}=-\frac{3}{x - 5}$. Then $-\frac{3(x + 3)}{(x - 5)(x + 3)}-\frac{11}{(x - 5)(x + 3)}=\frac{-3x-9 - 11}{(x - 5)(x + 3)}=\frac{-3x-20}{(x - 5)(x + 3)}$.

Step9: Simplify the ninth - fraction problem

For $\frac{1}{x + 1}-\frac{x}{x - 2}+\frac{x^{2}+2}{x^{2}-x - 2}$, factor $x^{2}-x - 2=(x + 1)(x - 2)$. Then $\frac{(x - 2)}{(x + 1)(x - 2)}-\frac{x(x + 1)}{(x + 1)(x - 2)}+\frac{x^{2}+2}{(x + 1)(x - 2)}=\frac{x - 2-x^{2}-x+x^{2}+2}{(x + 1)(x - 2)}=0$.

Answer:

1. $a + b$
2. $\frac{x}{x - 2}$
3. $\frac{2(3x + 4)}{5x - 2}$
4. $0$
5. $\frac{5x}{(x + 2)(x + 3)}$
6. $\frac{11y - 3}{(y + 3)^{2}(y - 3)}$
7. $\frac{p - 7}{p - 4}$
8. $\frac{-3x - 20}{(x - 5)(x + 3)}$
9. $0$