FRACCIONES ALGEBRAICAS

\[ \begin{aligned} &1) \quad \frac{2x}{4x} = \frac{1}{2} \\[10pt] &2) \quad \frac{3y}{9y} = \frac{1}{3} \\[10pt] &3) \quad \frac{5x^2}{10x} = \frac{x}{2} \\[10pt] &4) \quad \frac{7y^2}{14y} = \frac{y}{2} \end{aligned} \]

\[ \begin{aligned} &5) \quad \frac{8a}{16a^2} = \frac{1}{2a} \\[10pt] &6) \quad \frac{12x^3}{6x^2} = 2x \\[10pt] &7) \quad \frac{15b^2}{5b} = 3b \\[10pt] &8) \quad \frac{9x}{27x^3} = \frac{1}{3x^2} \end{aligned} \]

\[ \begin{aligned} &9) \quad \frac{6m^2}{18m} = \frac{m}{3} \\[10pt] &10) \quad \frac{10y^3}{5y^2} = 2y \\[10pt] &11) \quad \frac{3x+6}{x+2} = 3 \\[10pt] &12) \quad \frac{10y-5}{2y-1} = 5 \end{aligned} \]

\[ \begin{aligned} &13) \quad \frac{x^3 + 3x}{x} = x^2 + 3 \\[10pt] &14) \quad \frac{3x}{3x^2 + 6} = \frac{1}{x + 2} \\[10pt] &15) \quad \frac{m^2 – m}{m – 1} = m \\[10pt] &16) \quad \frac{2xy + 4xy}{5xy} = \frac{6}{5} \end{aligned} \]

\[ \begin{aligned} &17) \quad \frac{ab^2 + a^2b}{ab} = b + a \\[10pt] &18) \quad \frac{ab – ac}{b – c} = a \\[10pt] &19) \quad \frac{x^2 – x}{2x – 2} = \frac{x}{2} \\[10pt] &20) \quad \frac{6x^2 – 3x^2}{2x^2 – x} = \frac{3x}{2x – 1} \end{aligned} \]

\[ \begin{aligned} &21) \quad \frac{x – 1}{x^2 – 1} = \frac{1}{x + 1} \\[10pt] &22) \quad \frac{a^2 – 4}{a + 2} = a – 2 \\[10pt] &23) \quad \frac{m + 5}{m^2 – 25} = \frac{1}{m – 5} \\[10pt] &24) \quad \frac{x^2 – 9}{2x + 6} = \frac{x – 3}{2} \end{aligned} \]

\[ \begin{aligned} &25) \quad \frac{2x^2 – 2}{4x + 4} = \frac{x – 1}{2} \\[10pt] &26) \quad \frac{9x^2 – 1}{3x + 1} = 3x – 1 \\[10pt] &27) \quad \frac{4a^3b^3 – ab}{2ab + 1} = ab(2ab – 1) \\[10pt] &28) \quad \frac{9x – 3}{81x^2 – 9} = \frac{1}{3(3x + 1)} \end{aligned} \]

\[ \begin{aligned} &29) \quad \frac{2x^3 – 2x}{4x + 4} = \frac{x(x – 1)}{2} \\[10pt] &30) \quad \frac{2y^2 – 50}{y + 5} = 2(y – 5) \\[10pt] &31) \quad \frac{x^2 + 2x + 1}{x + 1} = x + 1 \\[10pt] &32) \quad \frac{x^2 + 4x + 4}{x + 2} = x + 2 \end{aligned} \]

\[ \begin{aligned} &33) \quad \frac{x^2 – 1}{x^2 + 2x + 1} = \frac{x – 1}{x + 1} \\[10pt] &34) \quad \frac{x^2 – 9}{x^2 + 6x + 9} = \frac{x – 3}{x + 3} \\[10pt] &35) \quad \frac{x^2 – 25}{x^2 – 10x + 25} = \frac{x + 5}{x – 5} \\[10pt] &36) \quad \frac{x^2 – 6x + 9}{x – 3} = x – 3 \end{aligned} \]

\[ \begin{aligned} &37) \quad \frac{2x + 3}{4x^2 + 12x + 9} = \frac{1}{2x + 3} \\[10pt] &38) \quad \frac{9x^2 – 6x + 1}{6x – 2} = \frac{3x – 1}{2} \\[10pt] &39) \quad \frac{25x^2 – 4}{25x^2 – 20x + 4} = \frac{5x + 2}{5x – 2} \\[10pt] &40) \quad \frac{16x^2 – 40x + 25}{16x^2 – 25} = \frac{4x – 5}{4x + 5} \end{aligned} \]

\[ \begin{aligned} &41) \quad \frac{x^3 – 1}{x^2 – 1} = \frac{x^2 + x + 1}{x + 1} \\[10pt] &42) \quad \frac{x^2 – 4}{x^3 – 8} = \frac{x + 2}{x^2 + 2x + 4} \\[10pt] &43) \quad \frac{x^2 – x + 1}{x^3 + 1} = \frac{1}{x + 1} \\[10pt] &44) \quad \frac{x^3 + 27}{x^2 + 6x + 9} = \frac{x^2 – 3x + 9}{x + 3} \end{aligned} \]

\[ \begin{aligned} &45) \quad \frac{4x^2 + 12x + 9}{8x^3 + 27} = \frac{2x + 3}{4x^2 – 6x + 9} \\[10pt] &46) \quad \frac{2x^2 + 6x + 18}{x^3 – 27} = \frac{2}{x – 3} \\[10pt] &47) \quad \frac{x^3 – 8}{x^2 – 2x} = \frac{x^2 + 2x + 4}{x} \\[10pt] &48) \quad \frac{x^3 – 27}{x^2 + 3x + 9} = x – 3 \end{aligned} \]

\[ \begin{aligned} &49) \quad \frac{x^2 + 2x + 4}{x^3 – 8} = \frac{1}{x – 2} \\[10pt] &50) \quad \frac{2x – 6}{10x^3 – 270} = \frac{1}{5(x^2 + 3x + 9)} \\[10pt] &51) \quad \frac{x^4 – 16}{x^2 – 4} = x^2 + 4 \\[10pt] &52) \quad \frac{x^2 – 9}{x^4 – 81} = \frac{1}{x^2 + 9} \end{aligned} \]

\[ \begin{aligned} &53) \quad \frac{x^6 – 1}{x^2 – 1} = x^4 + x^2 + 1 \\[10pt] &54) \quad \frac{x^3 + 3x^2 + 3x + 1}{x + 1} = (x + 1)^2 \\[10pt] &55) \quad \frac{16x^2 – 40x + 25}{16x^2 – 25} = \frac{4x – 5}{4x + 5} \\[10pt] &56) \quad \frac{x + 2y}{x^3 + 6x^2y + 12xy^2 + 8y^3} = \frac{1}{(x + 2y)^2} \end{aligned} \]

\[ \begin{aligned} &57) \quad \frac{27m^3 – 54m^2 + 36m – 8}{3m – 2} = 9m^2 – 12m + 4 \\[10pt] &58) \quad \frac{x^2 – 9y^2}{x^3 – 27y^3 – 9x^2y + 27xy^2} = \frac{x + 3y}{x^2 – 6xy + 9y^2} \\[10pt] &59) \quad \frac{p^3 + 6p^2 + 12p + 8}{p^3 + 8} = \frac{p^2 + 4p + 4}{p^2 – 2p + 4} \\[10pt] &60) \quad \frac{p^3 – 8}{4p^3 – 6p^2 + 12p – 8} = \frac{p^3 – 8}{2(2p^3 – 3p^2 + 6p – 4)} \end{aligned} \]

\[ \begin{aligned} &61) \quad \frac{x + 2}{x^2 + 5x + 6} \quad \text{(No se puede simplificar más)} \\[10pt] &62) \quad \frac{x^2 – 4}{x^2 – 5x + 6} = \frac{x + 2}{x – 3} \\[10pt] &63) \quad \frac{x^3 – 8}{x^2 – 5x + 6} = \frac{x^2 + 2x + 4}{x – 3} \\[10pt] &64) \quad \frac{x^3 – 8}{x^2 + 3x – 10} = \frac{x^2 + 2x + 4}{x + 5} \end{aligned} \]

\[ \begin{aligned} &65) \quad \frac{x^2 + 4x + 3}{x^2 + 3x + 2} = \frac{x + 3}{x + 2} \\[10pt] &66) \quad \frac{x^2 + x – 2}{x^2 – 2x + 1} = \frac{x + 2}{x – 1} \\[10pt] &67) \quad \frac{x^3 + 9x^2 + 27x + 27}{x^2 + 8x + 15} = \frac{x^2 + 6x + 9}{x + 5} \\[10pt] &68) \quad \frac{x^3 – 3x^2 + 3x – 1}{x^2 + 3x – 4} = \frac{x^2 – 2x + 1}{x + 4} \end{aligned} \]

\[ \begin{aligned} &69) \quad \frac{x^2 – x – 6}{x^3 + 6x^2 + 12x + 8} = \frac{x – 3}{x^2 + 4x + 4} \\[10pt] &70) \quad \frac{x^2 – 4x + 3}{x^3 – 9x^2 + 27x – 27} = \frac{x – 1}{x^2 – 6x + 9} \\[10pt] &71) \quad \frac{2x^2 + 3x + 1}{x^2 + 2x + 1} = \frac{2x + 1}{x + 1} \\[10pt] &72) \quad \frac{x^2 – 4x + 4}{3x^2 – 7x + 2} = \frac{x – 2}{3x – 1} \end{aligned} \]

\[ \begin{aligned} &73) \quad \frac{4x^2 + 11x – 3}{x^3 + 27} = \frac{4x^2 + 11x – 3}{(x + 3)(x^2 – 3x + 9)} \\[10pt] &74) \quad \frac{x^2 – 2x + 1}{7x^2 – 4x – 3} = \frac{x – 1}{7x + 3} \\[10pt] &75) \quad \frac{x^3 – 8}{3x^2 + x – 14} = \frac{x^2 + 2x + 4}{3(x + 7)} \\[10pt] &76) \quad \frac{\frac{1}{x^2} – x}{\frac{1}{x^2} + x} = \frac{1 – x^3}{1 + x^3} \end{aligned} \]

\[ \begin{aligned} &77) \quad \frac{m + \frac{1}{2}}{2 + \frac{1}{m}} = \frac{m^2 + m}{2m + 1} \\[10pt] &78) \quad \frac{\frac{1}{p} + \frac{1}{q}}{\frac{1}{p} – \frac{1}{q}} = \frac{q + p}{q – p} \\[10pt] &79) \quad \frac{\frac{1}{u^2} + \frac{1}{v^2}}{\frac{1}{u^2} – \frac{1}{v^2}} = \frac{v^2 + u^2}{v^2 – u^2} \\[10pt] &80) \quad \frac{1 + \frac{1}{\sqrt{x}}}{1 + \frac{1}{\sqrt{y}}} = \frac{\sqrt{x} + 1}{\sqrt{y} + 1} \\[10pt] \end{aligned} \]

\[ \begin{aligned} &81) \quad \frac{x + \frac{1}{x^2}}{x^2 + \frac{1}{x}} = \frac{x^3 + 1}{x^3 + 1} = 1 \\[10pt] &82) \quad \frac{1}{\frac{1}{p} + \frac{1}{q}} = \frac{pq}{p + q} \\[10pt] &83) \quad \frac{\frac{x}{x + 4}}{\frac{x + 5}{x}} = \frac{x^2}{(x + 4)(x + 5)} \\[10pt] &84) \quad \frac{\frac{x^2 + 2x + 1}{x}}{\frac{x^2 – 1}{x^2}} = \frac{x(x + 1)^2}{x^2 – 1} \\[10pt] \end{aligned} \]

\[ \begin{aligned} &85) \quad \frac{\frac{x^3 – 1}{x + 1}}{\frac{x – 1}{x^2 + 2x + 1}} = \frac{(x – 1)^2}{(x + 1)^2} \\[10pt] &86) \quad \frac{\frac{2x^2 – x – 1}{x^2 – 1}}{2x + 1} = \frac{2x – 1}{2x + 1} \\[10pt] &87) \quad \frac{\frac{1}{u^2} – \frac{1}{v^2}}{\frac{1}{v – u}} = \frac{v + u}{uv} \\[10pt] &88) \quad \frac{1 + t^{-3}}{1 – t^{-3}} = \frac{t^3 + 1}{t^3 – 1} \\[10pt] \end{aligned} \]

\[ \begin{aligned} &89) \quad \frac{\frac{r}{s} + 2}{\frac{s}{r} + 2} = \frac{r + 2s}{s + 2r} \\[10pt] &90) \quad \frac{\frac{a}{a – 1} – \frac{a + 1}{a}}{1 – \frac{a}{a – 1}} = \frac{1}{a} \\[10pt] &91) \quad \frac{5m^3}{15m^4} = \frac{1}{3m} \\[10pt] &92) \quad \frac{7p^3q^3 – 4p^2q^5}{p^2q^3} = 7p – 4q^2 \\[10pt] \end{aligned} \]

\[ \begin{aligned} &93) \quad \frac{m^2 – 2}{m + \sqrt{2}} = m – \sqrt{2} \\[10pt] &94) \quad \frac{x^2 – 11x + 30}{2x – 12} = \frac{x – 5}{2} \\[10pt] &95) \quad \frac{x^2 – 4}{x^3 – 8} = \frac{x + 2}{x^2 + 2x + 4} \\[10pt] &96) \quad \frac{x^2 + 2x + 1}{x^3 + 3x^2 + 3x + 1} = \frac{1}{x + 1} \\[10pt] \end{aligned} \]

\[ \begin{aligned} &97) \quad \frac{m^2 – 2}{m + \sqrt{2}} = m – \sqrt{2} \\[10pt] &98) \quad \frac{x^2 – 11x + 30}{2x – 12} = \frac{x – 5}{2} \\[10pt] &99) \quad \frac{x^2 – 4}{x^3 – 8} = \frac{x + 2}{x^2 + 2x + 4} \\[10pt] &100) \quad \frac{x^2 + 2x + 1}{x^3 + 3x^2 + 3x + 1} = \frac{1}{x + 1} \\[10pt] \end{aligned} \]