Reduce each of the following to the lowest terms:

$\displaystyle \text{Question 1: } \frac{x^2+x}{x^2-1}$

$\displaystyle \frac{x^2+x}{x^2-1} = \frac{x ( x+1 ) }{ ( x+1 ) ( x-1 ) } = \frac{x}{ ( x-1 ) }$

$\displaystyle \\$

$\displaystyle \text{Question 2: } \frac{x^2-16}{{ ( x+4 ) }^2}$

$\displaystyle \frac{x^2-16}{{ ( x+4 ) }^2} = \frac{ ( x-4 ) ( x+4 ) }{ ( x+4 ) ( x+4 ) } = \frac{x-4}{x+4}$

$\displaystyle \\$

$\displaystyle \text{Question 3: } \frac{a^2-b^2}{a^2b-{ab}^2}$

$\displaystyle \frac{a^2-b^2}{a^2b-{ab}^2} = \frac{ ( a-b ) (a+b)}{ab(a-b)} = \frac{(a+b)}{ab}$

$\displaystyle \\$

$\displaystyle \text{Question 4: } \frac{a^2+ab}{{2a}^3b-{2ab}^3}$

$\displaystyle \frac{a^2+ab}{{2a}^3b-{2ab}^3} = \frac{a ( a+b ) }{2ab ( a^2-b^2 ) } = \frac{a ( a+b ) }{2ab ( a-b ) ( a+b ) } = \frac{1}{2 b\ ( a-b ) }$

$\displaystyle \\$

$\displaystyle \text{Question 5: } \frac{x^2-9}{x^2+x-6}$

$\displaystyle \frac{x^2-9}{x^2+x-6} = \frac{ ( x-3 ) ( x+3 ) }{ ( x+3 ) ( x-2 ) } = \frac{ ( x-3 ) }{ ( x-2 ) }$

$\displaystyle \\$

$\displaystyle \text{Question 6: } \frac{x^2-2x}{x^2+3x-10}$

$\displaystyle \frac{x^2-2x}{x^2+3x-10} = \frac{x ( x-2 ) }{ ( x+5 ) ( x-2 ) } = \frac{x}{ ( x+5 ) }$

$\displaystyle \\$

$\displaystyle \text{Question 7: } \frac{x+3}{x^2-x-12}$

$\displaystyle \frac{x+3}{x^2-x-12} = \frac{x+3}{ ( x-4 ) ( x+3 ) } = \frac{1}{x-4}$

$\displaystyle \\$

$\displaystyle \text{Question 8: } \frac{x^2+2x-15}{x^2+4x-21}$

$\displaystyle \frac{x^2+2x-15}{x^2+4x-21} = \frac{ ( x+5 ) ( x-3 ) }{ ( x+7 ) ( x-3 ) } = \frac{x+5}{x+7}$

$\displaystyle \\$

$\displaystyle \text{Question 9: } \frac{x^2-5x+4}{x^2-3x-4}$

$\displaystyle \frac{x^2-5x+4}{x^2-3x-4} = \frac{ ( x-1 ) ( x-4 ) }{ ( x+1 ) ( x-4 ) } = \frac{x-1}{x+1}$

$\displaystyle \\$

$\displaystyle \text{Question 10: } \frac{x^3-{xy}^2}{x^3+{2x}^2y+{xy}^2}$

$\displaystyle \frac{x^3-{xy}^2}{x^3+{2x}^2y+{xy}^2} = \frac{x ( x-y ) ( x+y ) }{x^2 ( x+y ) +2y ( x+y ) } = \frac{x ( x-y ) ( x+y ) }{x ( x+y ) ( x+y ) } = \frac{x-y}{x+y}$

$\displaystyle \\$

$\displaystyle \text{Question 11: } \frac{{2x}^2+7x+3}{{3x}^2+10x+3}$

$\displaystyle \frac{{2x}^2+7x+3}{{3x}^2+10x+3} = \frac{{2x}^2+x+6x+3}{3x^2+9x+x+3}$

$\displaystyle = \frac{x ( 2x+1 ) +3 ( 2x+1 ) }{3x ( x+3 ) + ( x+3 ) } = \frac{ ( 2x+1 ) (x+3)}{ ( 3x+1 ) (x+3)} = \frac{2x+1}{3x+1}$

$\displaystyle \\$

$\displaystyle \text{Question 12: } \frac{{2x}^2+x-3}{{3x}^2+x-4}$

$\displaystyle \frac{{2x}^2+x-3}{{3x}^2+x-4} = \frac{2x^2-2x+3x-3}{3x^2+4x-3x-4}$

$\displaystyle = \frac{2x ( x-1 ) +3 ( x-1 ) }{3x ( x-1 ) +4 ( x-1 ) } = \frac{ ( 2x+3 ) ( x-1 ) }{ ( 3x+4 ) ( x-1 ) } = \frac{2x+3}{3x+4}$

$\displaystyle \\$

Simplify:

$\displaystyle \text{Question 13: } \frac{x^2-16}{x^2-9} \times \frac{x+3}{x+4}$

$\displaystyle \frac{x^2-16}{x^2-9} \times \frac{x+3}{x+4} = \frac{ ( x-4 ) ( x+4 ) ( x+3 ) }{ ( x-3 ) ( x+3 ) ( x+4 ) } = \frac{x-4}{x-3}$

$\displaystyle \\$

$\displaystyle \text{Question 14: } \frac{a^2-{ab}^2}{a^2-{16b}^2} \times \frac{a+4b}{a-3b}$

$\displaystyle \frac{a^2-{ab}^2}{a^2-{16b}^2} \times \frac{a+4b}{a-3b} = \frac{ ( a-3b ) (a+3b(a+4b)}{ ( a-4b ) ( a+4b ) ( a-3b ) } = \frac{a+3b}{a-4b}$

$\displaystyle \\$

$\displaystyle \text{Question 15: } \frac{x^2+6x+5}{x^2+5x} \times \frac{x^3-x}{x^2-1}$

$\displaystyle \frac{x^2+6x+5}{x^2+5x} \times \frac{x^3-x}{x^2-1} = \frac{ ( x+1 ) ( x+5 ) x ( x-1 ) ( x+1 ) }{x ( x+5 ) ( x-1 ) ( x+1 ) } = (x+1)$

$\displaystyle \\$

$\displaystyle \text{Question 16: } \frac{x+y}{x^2-xy} \div \frac{x^2+xy}{x-y}$

$\displaystyle \frac{x+y}{x^2-xy} \div \frac{x^2+xy}{x-y} = \frac{x+y}{x(x-y)} \times \frac{x-y}{x(x+y)} = \frac{1}{x^2}$

$\displaystyle \\$

$\displaystyle \text{Question 17: } \frac{x^2-5x}{2x-3y} \div \frac{x^2-25}{{4x}^2-9y^2}$

$\displaystyle \frac{x^2-5x}{2x-3y} \div \frac{x^2-25}{{4x}^2-9y^2} = \frac{x ( x-5 ) }{ ( 2x-3y ) } \times \frac{ ( 2x-3y ) ( 2x+3y ) }{ ( x-5 ) ( x+5 ) } = \frac{x(2x+3y)}{(x+5)}$

$\displaystyle \\$

$\displaystyle \text{Question 18: } \frac{x^2+2-6}{x^2+2x-3} \div \frac{x^2+5x-14}{x^2+4x-5}$

$\displaystyle \frac{x^2+2-6}{x^2+2x-3} \div \frac{x^2+5x-14}{x^2+4x-5} = \frac{ ( x+3 ) ( x-2 ) }{ ( x+3 ) ( x-1 ) } \times \frac{ ( x+5 ) ( x-1 ) }{ ( x+7 ) ( x-2 ) } = \frac{x+5}{x+7}$

$\displaystyle \\$

$\displaystyle \text{Question 19: } \frac{{3x}^2-x-2}{x^2+x-2} \div \frac{{3x}^2-7x-6}{x^2-x-6}$

$\displaystyle \frac{{3x}^2-x-2}{x^2+x-2} \div \frac{{3x}^2-7x-6}{x^2-x-6}$

$\displaystyle = \frac{{3x}^2-3x+2x-2}{x^2+2x-x-2} \times \frac{x^2-3x+2x-6}{3x^2-9x+2x-6}$

$\displaystyle = \frac{3x ( x-1 ) +2 ( x-1 ) }{x ( x+2 ) -1 ( x+2 ) } \times \frac{x ( x+2 ) -3 ( x+2 ) }{3x ( x-3 ) +2 ( x-3 ) }$

$\displaystyle = \frac{ ( 3x+2 ) ( x-1 ) }{ ( x-1 ) ( x+2 ) } \times \frac{ ( x+2 ) ( x-3 ) }{ ( x-3 ) ( 3x+2 ) } = 1$

$\displaystyle \\$

$\displaystyle \text{Question 20: } \frac{{2x}^2-3x-2}{x^2+7x+12} \times \frac{x^2+x-12}{x^2+3x-10} \div \frac{{2x}^2-5x-3}{x^2+8x+15}$

$\displaystyle \frac{{2x}^2-3x-2}{x^2+7x+12} \times \frac{x^2+x-12}{x^2+3x-10} \div \frac{{2x}^2-5x-3}{x^2+8x+15}$

$\displaystyle = \frac{{2x}^2-4x+x-2}{ ( x+4 ) ( x+3 ) } \times \frac{ ( x+4 ) ( x-3 ) }{ ( x+5 ) ( x-2 ) } \times \frac{ ( x+3 ) ( x+5 ) }{{2x}^2-2x+3x-3}$

$\displaystyle = \frac{2x ( x-2 ) + ( x-2 ) }{ ( x+4 ) ( x+3 ) } \times \frac{ ( x+4 ) ( x-3 ) }{ ( x+5 ) ( x-2 ) } \times \frac{ ( x+3 ) ( x+5 ) }{ ( 2x+1 ) ( x-3 ) }$

$\displaystyle = \frac{(2x+1) ( x-2 ) }{ ( x+4 ) ( x+3 ) } \times \frac{ ( x+4 ) ( x-3 ) }{ ( x+5 ) ( x-2 ) } \times \frac{ ( x+3 ) ( x+5 ) }{ ( 2x+1 ) ( x-3 ) } = 1$

$\displaystyle \\$

Simplify:

$\displaystyle \text{Question 21: } \frac{2x+3}{3}-\frac{2x-4}{4}$

$\displaystyle \frac{2x+3}{3}-\frac{2x-4}{4}$

$\displaystyle = \frac{4 ( 2x+3 ) -3 ( 2x-4 ) }{12}$

$\displaystyle = \frac{8x+12-6x+12}{12} = \frac{2x+24}{12} = \frac{x+12}{6}$

$\displaystyle = \frac{3}{x-1}-\frac{3}{x+1}$

$\displaystyle = \frac{3 ( x+1 ) -3 ( x-1 ) }{ ( x-1 ) ( x+1 ) }$

$\displaystyle = \frac{3x+3-3x+3}{ ( x-1 ) (x+1)} = \frac{6}{ ( x-1 ) (x+1)}$

$\displaystyle \\$

$\displaystyle \text{Question 22: } \frac{3x-1}{4x}+\frac{3-5x}{12x}$

$\displaystyle \frac{3x-1}{4x}+\frac{3-5x}{12x}$

$\displaystyle = \frac{12x ( 3x-1 ) +4x ( 3-5x ) }{48x^2}$

$\displaystyle = \frac{3}{x-1}-\frac{3}{x+1}$

$\displaystyle = \frac{3 ( x+1 ) -3 ( x-1 ) }{ ( x-1 ) ( x+1 ) }$

$\displaystyle = \frac{3x+3-3x+3}{ ( x-1 ) (x+1)} = \frac{6}{ ( x-1 ) (x+1)}$

$\displaystyle \\$

$\displaystyle \text{Question 23: } \frac{3x-1}{4x}+\frac{3-5x}{12x}$

$\displaystyle \frac{3x-1}{4x}+\frac{3-5x}{12x}$

$\displaystyle = \frac{12x ( 3x-1 ) +4x ( 3-5x ) }{48x^2}$

$\displaystyle = \frac{36x^2-12x+12x-20x^2}{48x^2} = \frac{{16x}^2}{48x^2} = \frac{1}{3}$

$\displaystyle \\$

$\displaystyle \text{Question 24: } \frac{x}{x-1}-\frac{x^2}{x^2-1}$

$\displaystyle \frac{x}{x-1}-\frac{x^2}{x^2-1} = \frac{x ( x^2-1 ) -x^2 ( x-1 ) }{ ( x-1 ) ( x-1 ) ( x+1 ) }$

$\displaystyle = \frac{x^3-x-x^3++x^2}{ ( x-1 ) ( x-1 ) ( x+1 ) } = \frac{x ( x-1 ) }{ ( x-1 ) ( x-1 ) ( x+1 ) }$

$\displaystyle = \frac{x}{ ( x-1 ) (x+1)}$

$\displaystyle \\$

$\displaystyle \text{Question 25: } \frac{6}{a+b}-\frac{5}{a-b}+\frac{11b}{a^2-b^2}$

$\displaystyle \frac{6}{a+b}-\frac{5}{a-b}+\frac{11b}{a^2-b^2}$

$\displaystyle = \frac{6 ( a-b ) -5 ( a+b ) +11b}{ ( a-b ) ( a+b ) }$

$\displaystyle = \frac{6a-6b-5a-5b+11b}{ ( a-b ) (a+b)}$

$\displaystyle = \frac{a}{ ( a-b ) ( a+b ) }$

$\displaystyle \\$

$\displaystyle \text{Question 26: } \frac{6}{x-2}+\frac{8}{2x-4}$

$\displaystyle \frac{6}{x-2}+\frac{8}{2x-4}$

$\displaystyle = \frac{6 ( 2x-4 ) +8 ( x-2 ) }{ ( x-2 ) ( 2x-4 ) } = \frac{12x-24+8x-16}{ ( x-2 ) ( 2x-4 ) }$

$\displaystyle = \frac{20x-40}{ ( x-2 ) ( 4-4 ) } = \frac{20 ( x-2 ) }{ ( x-2 ) ( x-4 ) } = \frac{20}{x-4}$

$\displaystyle \\$

$\displaystyle \text{Question 27: } \frac{1}{x^2-11x+30}+\frac{1}{x^2-9x+20}$

$\displaystyle \frac{1}{x^2-11x+30}+\frac{1}{x^2-9x+20}$

$\displaystyle = \frac{1}{ ( x-6 ) ( x-5 ) }+\frac{1}{ ( x-5 ) ( x-4 ) }$

$\displaystyle = \frac{ ( x-4 ) + ( x-6 ) }{ ( x-6 ) ( x-5 ) ( x-4 ) }$

$\displaystyle = \frac{2(x-5)}{ ( x-6 ) ( x-5 ) ( x-4 ) } = \frac{2}{ ( x-6 ) (x-4)}$

$\displaystyle \\$

$\displaystyle \text{Question 28: } \frac{1}{x^2-8x+15}+\frac{1}{x^2-4x+3}-\frac{2}{ ( x^2-6x+5 ) }$

$\displaystyle \frac{1}{x^2-8x+15}+\frac{1}{x^2-4x+3}-\frac{2}{ ( x^2-6x+5 ) }$

$\displaystyle = \frac{1}{ ( x-3 ) ( x-5 ) }+\frac{1}{ ( x-3 ) ( x-1 ) }-\frac{2}{ ( x-3 ) ( x-2 ) }$

$\displaystyle = \frac{ ( x-1 ) ( x-2 ) + ( x-2 ) ( x-5 ) -2 ( x-1 ) ( x-5 ) }{ ( x-1 ) ( x-2 ) ( x-3 ) ( x-5 ) }$

$\displaystyle = \frac{x^2-3x+2+x^2-7x+10-2x^2+12x-10}{ ( x-1 ) ( x-2 ) ( x-3 ) ( x-5 ) }$

$\displaystyle = \frac{2(x+1)}{ ( x-1 ) ( x-2 ) ( x-3 ) (x-5)}$

$\displaystyle \\$

$\displaystyle \text{Question 29: } \frac{x-3}{x^2-7x+12}+\frac{2 ( x-1 ) }{x^2-4x+3}-\frac{3 ( x-4 ) }{x^2-5x+4}$

$\displaystyle \frac{x-3}{x^2-7x+12}+\frac{2 ( x-1 ) }{x^2-4x+3}-\frac{3 ( x-4 ) }{x^2-5x+4}$

$\displaystyle = \frac{x-3}{ ( x-3 ) ( x-4 ) }+\frac{2 ( x-1 ) }{ ( x-3 ) ( x-1 ) }-\frac{3 ( x-4 ) }{ ( x-4 ) ( x-1 ) }$

$\displaystyle = \frac{1}{ ( x-4 ) }+\frac{2}{ ( x-3 ) }-\frac{3}{ ( x-1 ) }$

$\displaystyle = \frac{x^2-4x+3+2x^2-10x+8-{3x}^2+21x-36}{ ( x-1 ) ( x-3 ) ( x-4 ) }$

$\displaystyle = \frac{7x-25}{ ( x-1 ) ( x-3 ) (x-4)}$

$\displaystyle \\$

$\displaystyle \text{Question 30: } \frac{x^2-x-6}{x^2-9}+\frac{x^2+2x-24}{x^2-x-12}$

$\displaystyle \frac{x^2-x-6}{x^2-9}+\frac{x^2+2x-24}{x^2-x-12}$

$\displaystyle = \frac{ ( x-6 ) (x+1)}{ ( x-3 ) (x+3)}+\frac{ ( x+6 ) (x-4)}{ ( x-4 ) (x+3)}$

$\displaystyle = \frac{ ( x-6 ) ( x+1 ) }{ ( x-3 ) ( x+3 ) }+\frac{ ( x+6 ) }{ ( x+3 ) }$

$\displaystyle = \frac{(x^2-5x-6)+(x^2+3x-18)}{ ( x-3 ) ( x+3 ) }$

$\displaystyle = \frac{{2x}^2-2x-24}{ ( x-3 ) (x+3)} = \frac{2 ( x-4 ) (x+3)}{ ( x-3 ) (x+3)} = \frac{2(x-4)}{(x-3)}$

$\displaystyle \\$

$\displaystyle \text{Question 31: }( \frac{2x}{4x+1}-\frac{x}{2x+3} ) \div{}\ \frac{10x+5}{2x+3}$

$\displaystyle ( \frac{2x}{4x+1}-\frac{x}{2x+3} ) \div{}\ \frac{10x+5}{2x+3}$

$\displaystyle = \frac{4x^2+6x-{4x}^2-x}{ ( 4x+1 ) ( 2x+3 ) }\times{}\frac{2x+3}{5 ( 2x+1 ) }$

$\displaystyle = \frac{5x(2x+3)}{5 ( 4x+1 ) ( 2x+3 ) (2x+1)} = \frac{x}{ ( 4x+1 ) (2x+1)}$

$\displaystyle \\$

$\displaystyle \text{Question 32: }( \frac{a^2+a-12}{a^2+6a+8}+\frac{a^2-6a+5}{a^2-3a-10} )$ of $\displaystyle \frac{a^2+8+12}{a^2+4a-12}$

$\displaystyle ( \frac{a^2+a-12}{a^2+6a+8}+\frac{a^2-6a+5}{a^2-3a-10} )$ of $\displaystyle \frac{a^2+8+12}{a^2+4a-12}$
$\displaystyle = ( \frac{ ( a+4 ) ( a+3 ) }{ ( a+4 ) ( a+2 ) }+\frac{ ( a-5 ) ( a-1 ) }{ ( a-5 ) ( a+2 ) } ) \times \frac{(a+2)(a+6)}{(a+6)(a-2)}$
$\displaystyle = ( \frac{a-3}{a+2}+\frac{a-1}{a+2} ) = \ \frac{2a-4}{a-2}$