\(\textbf{1)}\) \( \displaystyle\frac{5}{x}+\frac{x^2}{7} \)
The answer is \( \displaystyle\frac{x^3+35}{7x} \)
\(\,\,\,\,\,\displaystyle\frac{5}{x}+\frac{x^2}{7}\)
\(\,\,\,\,\,\displaystyle\frac{7}{7} \cdot \frac{5}{x}+\frac{x^2}{7} \cdot \frac{x}{x}\)
\(\,\,\,\,\,\displaystyle\frac{35}{7x}+\frac{x^3}{7x}\)
\(\,\,\,\,\,\displaystyle\frac{x^3+35}{7x}\)
\(\textbf{2)}\) \( \displaystyle\frac{x-5}{x+3}+\frac{x+2}{x^2+5x+6} \)
The answer is \( \displaystyle\frac{x-4}{x+3} \)
\(\,\,\,\,\,\displaystyle\frac{x-5}{x+3}+\frac{x+2}{x^2+5x+6}\)
\(\,\,\,\,\,\displaystyle\frac{x-5}{x+3}+\frac{x+2}{\left(x+2\right)\left(x+3\right)}\)
\(\,\,\,\,\,\displaystyle \frac{x+2}{x+2} \cdot \frac{x-5}{x+3}+\frac{x+2}{\left(x+2\right)\left(x+3\right)}\)
\(\,\,\,\,\,\displaystyle \frac{\left(x+2\right)\left(x-5\right)}{\left(x+2\right)\left(x+3\right)}+\frac{x+2}{\left(x+2\right)\left(x+3\right)}\)
\(\,\,\,\,\,\displaystyle \frac{x^2+5x+6}{\left(x+2\right)\left(x+3\right)}+\frac{x+2}{\left(x+2\right)\left(x+3\right)}\)
\(\,\,\,\,\,\displaystyle \frac{x^2+6x+8}{\left(x+2\right)\left(x+3\right)}\)
\(\,\,\,\,\,\displaystyle \frac{\left(x+2\right)\left(x-4\right)}{\left(x+2\right)\left(x+3\right)}\)
\(\,\,\,\,\,\displaystyle\frac{x-4}{x+3} \)
\(\textbf{3)}\) \( \displaystyle\frac{5}{x+3}+\frac{x}{x^2-9} \)
The answer is \( \displaystyle\frac{3(2x-5)}{(x-3)(x+3)} \)
\(\,\,\,\,\,\displaystyle\frac{5}{x+3}+\frac{x}{x^2-9}\)
\(\,\,\,\,\,\displaystyle\frac{5}{x+3}+\frac{x}{\left(x-3\right)\left(x+3\right)}\)
\(\,\,\,\,\,\displaystyle\frac{x-3}{x-3} \cdot \frac{5}{x+3}+\frac{x}{\left(x-3\right)\left(x+3\right)}\)
\(\,\,\,\,\,\displaystyle\frac{5x-15}{\left(x-3\right)\left(x+3\right)}+\frac{x}{\left(x-3\right)\left(x+3\right)}\)
\(\,\,\,\,\,\displaystyle\frac{6x-15}{\left(x-3\right)\left(x+3\right)}\)
\(\,\,\,\,\,\displaystyle\frac{3(2x-5)}{(x-3)(x+3)} \)
\(\textbf{4)}\) \( \displaystyle\frac{5}{x-2}-\frac{8}{2-x} \)
The answer is \( \displaystyle\frac{13}{x-2} \)
\(\,\,\,\,\,\displaystyle\frac{5}{x-2}-\frac{8}{2-x}\)
\(\,\,\,\,\,\displaystyle\frac{5}{x-2}-\frac{8}{-\left(x-2\right)}\)
\(\,\,\,\,\,\displaystyle\frac{5}{x-2}+\frac{8}{\left(x-2\right)}\)
\(\,\,\,\,\,\displaystyle\frac{13}{x-2} \)
\(\textbf{5)}\) \( \displaystyle\frac{1}{2x}+\frac{x}{3x}+\frac{3}{4x} \)
The answer is \( \displaystyle\frac{4x+15}{12x} \)
\(\,\,\,\,\,\displaystyle\frac{1}{2x}+\frac{x}{3x}+\frac{3}{4x}\)
\(\,\,\,\,\,\displaystyle\frac{1}{2x} \cdot\frac{6}{6} +\frac{x}{3x} \cdot\frac{4}{4} +\frac{3}{4x} \cdot\frac{3}{3} \)
\(\,\,\,\,\,\displaystyle\frac{6}{12x} +\frac{4x}{12x} +\frac{9}{12x} \)
\(\,\,\,\,\,\displaystyle\frac{6+4x+9}{12x}\)
\(\,\,\,\,\,\displaystyle\frac{4x+15}{12x} \)
\(\textbf{6)}\) \( \displaystyle\frac{5}{x+3}+4 \)
The answer is \( \displaystyle\frac{4x+17}{x+3} \)
\(\,\,\,\,\,\displaystyle\frac{5}{x+3}+4\)
\(\,\,\,\,\,\displaystyle\frac{5}{x+3}+\frac{4}{1} \cdot \frac{x+3}{x+3}\)
\(\,\,\,\,\,\displaystyle\frac{5}{x+3}+\frac{4x+12}{x+3}\)
\(\,\,\,\,\,\displaystyle\frac{4x+17}{x+3} \)
\(\textbf{7)}\) \( \displaystyle\frac{x}{x+5}+\frac{3x+1}{x+2} \)
The answer is \( \displaystyle\frac{4x^2+18x+5}{(x+2)(x+5)} \)
\(\,\,\,\,\,\displaystyle\frac{x}{x+5}+\frac{3x+1}{x+2}\)
\(\,\,\,\,\,\displaystyle\frac{x+2}{x+2} \cdot \frac{x}{x+5}+\frac{3x+1}{x+2} \cdot \frac{x+5}{x+5}\)
\(\,\,\,\,\,\displaystyle\frac{x^2+2x}{\left(x+2\right)\left(x+5\right)}+\frac{\left(3x+1\right)\left(x+5\right)}{\left(x+2\right)\left(x+5\right)}\)
\(\,\,\,\,\,\displaystyle\frac{x^2+2x}{\left(x+2\right)\left(x+5\right)}+\frac{3x^2+16x+5}{\left(x+2\right)\left(x+5\right)}\)
\(\,\,\,\,\,\displaystyle\frac{4x^2+18x+5}{\left(x+2\right)\left(x+5\right)}\)
\(\textbf{8)}\) \( \displaystyle\frac{2x+5}{x^2+3x+2}+\frac{x-1}{x^2+7x+10} \)
The answer is \( \displaystyle\frac{3x^2+15x+24}{\left(x+5\right)\left(x+1\right)\left(x+2\right)} \)
\(\,\,\,\,\,\displaystyle\frac{2x+5}{x^2+3x+2}+\frac{x-1}{x^2+7x+10}\)
\(\,\,\,\,\,\displaystyle\frac{2x+5}{\left(x+1\right)\left(x+2\right)}+\frac{x-1}{\left(x+5\right)\left(x+2\right)}\)
\(\,\,\,\,\,\displaystyle\frac{x+5}{x+5} \cdot \frac{2x+5}{\left(x+1\right)\left(x+2\right)}+\frac{x-1}{\left(x+5\right)\left(x+2\right)} \cdot \frac{x+1}{x+1}\)
\(\,\,\,\,\,\displaystyle\frac{\left(x+5\right)\left(2x+5\right)}{\left(x+5\right)\left(x+1\right)\left(x+2\right)}+\frac{\left(x-1\right)\left(x+1\right)}{\left(x+5\right)\left(x+2\right)\left(x+1\right)}\)
\(\,\,\,\,\,\displaystyle\frac{2x^2+15x+25}{\left(x+5\right)\left(x+1\right)\left(x+2\right)}+\frac{x^2-1}{\left(x+5\right)\left(x+2\right)\left(x+1\right)}\)
\(\,\,\,\,\,\displaystyle\frac{3x^2+15x+24}{\left(x+5\right)\left(x+1\right)\left(x+2\right)}\)
\(\,\,\,\,\,\displaystyle\frac{3x^2+15x+24}{\left(x+5\right)\left(x+1\right)\left(x+2\right)}\)
\(\textbf{9)}\) \( \displaystyle\frac{x+7}{x^2+4x+4}+\frac{x+1}{x^2-4} \)
The answer is \( \displaystyle\frac{2(x-8)}{(x-2)(x+2)^2} \)
\(\,\,\,\,\,\displaystyle\frac{x+7}{x^2+4x+4}+\frac{x+1}{x^2-4}\)
\(\,\,\,\,\,\displaystyle\frac{x+7}{\left(x+2\right)\left(x+2\right)}+\frac{x+1}{\left(x+2\right)\left(x-2\right)}\)
\(\,\,\,\,\,\displaystyle\frac{x-2}{x-2} \cdot \frac{x+7}{\left(x+2\right)\left(x+2\right)}+\frac{x+1}{\left(x+2\right)\left(x-2\right)} \cdot \frac{x+2}{x+2}\)
\(\,\,\,\,\,\displaystyle\frac{x^2+5x-14}{\left(x-2\right)\left(x+2\right)^2}+\frac{x^2+3x+2}{\left(x-2\right)\left(x+2\right)^2}\)
\(\,\,\,\,\,\displaystyle\frac{2x^2+8x-12}{\left(x-2\right)\left(x+2\right)^2}+\frac{x^2+3x+2}{\left(x-2\right)\left(x+2\right)^2}\)
\(\textbf{10)}\) \( \displaystyle\frac{2x}{x+2}+\frac{4}{x+2} \)
The answer is \( 2 \)
\(\,\,\,\,\,\displaystyle\frac{2x}{x+2}+\frac{4}{x+2}\)
\(\,\,\,\,\,\displaystyle\frac{2x+4}{x+2}\)
\(\,\,\,\,\,\displaystyle\frac{2\left(x+2\right)}{x+2}\)
\(\,\,\,\,\,2 \)
\(\textbf{11)}\) \( \displaystyle\frac{3x+1}{\left(x+2\right)^2}+\frac{3}{x+2} \)
The answer is \( \displaystyle\frac{6x+7}{(x+2)^2} \)
\(\,\,\,\,\,\displaystyle\frac{3x+1}{(x+2)^2}+\frac{3}{x+2}\)
\(\,\,\,\,\,\displaystyle\frac{3x+1}{(x+2)^2}+\frac{3}{x+2} \cdot \frac{x+2}{x+2}\)
\(\,\,\,\,\,\displaystyle\frac{3x+1}{\left(x+2\right)^2}+\frac{3x+6}{\left(x+2\right)^2}\)
\(\,\,\,\,\,\displaystyle\frac{6x+7}{(x+2)^2} \)
\(\textbf{12)}\) \( \displaystyle\frac{4}{x}+\frac{3}{x+1} \)
The answer is \( \displaystyle\frac{7x+4}{x(x+1)} \)
\(\,\,\,\,\,\displaystyle\frac{4}{x}+\frac{3}{x+1}\)
\(\,\,\,\,\,\displaystyle\frac{x+1}{x+1} \cdot \frac{4}{x}+\frac{3}{x+1} \cdot \frac{x}{x}\)
\(\,\,\,\,\,\displaystyle\frac{4x+4}{x\left(x+1\right)}+\frac{3x}{x\left(x+1\right)}\)
\(\,\,\,\,\,\displaystyle\frac{7x+4}{x(x+1)} \)
\(\textbf{13)}\) \( \displaystyle\frac{7}{x} – \frac{x^2}{5} \)
The answer is \( \displaystyle\frac{35 – x^3}{5x} \)
\(\,\,\,\,\,\displaystyle\frac{7}{x} – \frac{x^2}{5}\)
\(\,\,\,\,\,\displaystyle\frac{5}{5} \cdot \frac{7}{x} – \frac{x^2}{5} \cdot \frac{x}{x}\)
\(\,\,\,\,\,\displaystyle\frac{35}{5x} – \frac{x^3}{5x}\)
\(\,\,\,\,\,\displaystyle\frac{35 – x^3}{5x}\)
\(\textbf{14)}\) \( \displaystyle\frac{x+4}{x+1} – \frac{x-3}{x^2+2x+1} \)
The answer is \( \displaystyle\frac{x^2+4x+7}{(x+1)^2} \)
\(\,\,\,\,\,\displaystyle\frac{x+4}{x+1} – \frac{x-3}{x^2+2x+1}\)
\(\,\,\,\,\,\displaystyle\frac{x+4}{x+1} – \frac{x-3}{(x+1)(x+1)}\)
\(\,\,\,\,\,\displaystyle\frac{x+1}{x+1} \cdot \frac{x+4}{x+1} – \frac{x-3}{(x+1)(x+1)}\)
\(\,\,\,\,\,\displaystyle\frac{(x+4)(x+1) – (x-3)}{(x+1)^2}\)
\(\,\,\,\,\,\displaystyle\frac{x^2+5x+4 – x + 3}{(x+1)^2}\)
\(\,\,\,\,\,\displaystyle\frac{x^2+4x+7}{(x+1)^2}\)
\(\textbf{15)}\) \( \displaystyle\frac{4}{x-2} – \frac{6}{2-x} \)
The answer is \( \displaystyle\frac{10}{x-2} \)
\(\,\,\,\,\,\displaystyle\frac{4}{x-2} – \frac{6}{2-x}\)
\(\,\,\,\,\,\displaystyle\frac{4}{x-2} – \frac{6}{-(x-2)}\)
\(\,\,\,\,\,\displaystyle\frac{4}{x-2} + \frac{6}{x-2}\)
\(\,\,\,\,\,\displaystyle\frac{10}{x-2}\)
\(\textbf{16)}\) \( \displaystyle\frac{3x}{x+5} – \frac{2x+7}{x+1} \)
The answer is \( \displaystyle\frac{x^2 – 14x – 35}{(x+5)(x+1)} \)
\(\,\,\,\,\,\displaystyle\frac{3x}{x+5} – \frac{2x+7}{x+1}\)
\(\,\,\,\,\,\displaystyle\frac{x+1}{x+1} \cdot \frac{3x}{x+5} – \frac{2x+7}{x+1} \cdot \frac{x+5}{x+5}\)
\(\,\,\,\,\,\displaystyle\frac{3x(x+1)}{(x+5)(x+1)} – \frac{(2x+7)(x+5)}{(x+5)(x+1)}\)
\(\,\,\,\,\,\displaystyle\frac{3x^2+3x – (2x^2+17x+35)}{(x+5)(x+1)}\)
\(\,\,\,\,\,\displaystyle\frac{x^2 – 14x – 35}{(x+5)(x+1)}\)
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