Inicio › Matemáticas 2 › Jerarquía con fracciones

Matemáticas 2 · Tema 2

Jerarquía con fracciones

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Ejemplos

Ejemplo 1
13+45−63+36−65=−53−25+36=−50−12+1530=−4730\begin{aligned} \frac{1}{3} + \frac{4}{5} - \frac{6}{3} + \frac{3}{6} - \frac{6}{5} &= \\[14pt] -\frac{5}{3} - \frac{2}{5} + \frac{3}{6} &= \\[14pt] \frac{-50 - 12 + 15}{30} &= \mathbf{-\frac{47}{30}} \end{aligned}
Ejemplo 2
(35)2−2(13−45)−32=925−21(5−1215)−32=925−21(−715)−32=925+1415−32=54+140−225150=−31150\begin{aligned} \left(\frac{3}{5}\right)^2 - 2\left(\frac{1}{3} - \frac{4}{5}\right) - \frac{3}{2} &= \\[14pt] \frac{9}{25} - \frac{2}{1}\left(\frac{5-12}{15}\right) - \frac{3}{2} &= \\[14pt] \frac{9}{25} - \frac{2}{1}\left(-\frac{7}{15}\right) - \frac{3}{2} &= \\[14pt] \frac{9}{25} + \frac{14}{15} - \frac{3}{2} &= \\[14pt] \frac{54 + 140 - 225}{150} &= \mathbf{-\frac{31}{150}} \end{aligned}
Ejemplo 3
12[34−5(2−13)]=12[34−51(53)]=12[34−253]=12[9−10012]=−9124\begin{aligned} \frac{1}{2}\left[\frac{3}{4} - 5\left(2 - \frac{1}{3}\right)\right] &= \\[14pt] \frac{1}{2}\left[\frac{3}{4} - \frac{5}{1}\left(\frac{5}{3}\right)\right] &= \\[14pt] \frac{1}{2}\left[\frac{3}{4} - \frac{25}{3}\right] &= \\[14pt] \frac{1}{2}\left[\frac{9 - 100}{12}\right] &= \mathbf{-\frac{91}{24}} \end{aligned}
Ejemplo 4
32[14−3]225−3(14)=32[1−124]225−34=32[−114]28−1520=32[12116]−720=36332−720=−181556\begin{aligned} \frac{\dfrac{3}{2}\left[\dfrac{1}{4}-3\right]^2}{\dfrac{2}{5}-3\left(\dfrac{1}{4}\right)} &= \\[14pt] \frac{\dfrac{3}{2}\left[\dfrac{1-12}{4}\right]^2}{\dfrac{2}{5}-\dfrac{3}{4}} &= \\[14pt] \frac{\dfrac{3}{2}\left[\dfrac{-11}{4}\right]^2}{\dfrac{8-15}{20}} &= \\[14pt] \frac{\dfrac{3}{2}\left[\dfrac{121}{16}\right]}{-\dfrac{7}{20}} = \frac{\dfrac{363}{32}}{-\dfrac{7}{20}} &= \mathbf{-\frac{1815}{56}} \end{aligned}

Ejercicios

Bloque 1
14−65+64+35−710=\dfrac{1}{4}-\dfrac{6}{5}+\dfrac{6}{4}+\dfrac{3}{5}-\dfrac{7}{10}=
(12)3−4(32−56)−14=\left(\dfrac{1}{2}\right)^3-4\left(\dfrac{3}{2}-\dfrac{5}{6}\right)-\dfrac{1}{4}=
23[12+4(3−25)]=\dfrac{2}{3}\left[\dfrac{1}{2}+4\left(3-\dfrac{2}{5}\right)\right]=
35[12−4]232−2(16)=\dfrac{\dfrac{3}{5}\left[\dfrac{1}{2}-4\right]^2}{\dfrac{3}{2}-2\left(\dfrac{1}{6}\right)}=
Bloque 2
−24+63−84−73+16=-\dfrac{2}{4}+\dfrac{6}{3}-\dfrac{8}{4}-\dfrac{7}{3}+\dfrac{1}{6}=
(25)2−35−4(12−3)=\left(\dfrac{2}{5}\right)^2-\dfrac{3}{5}-4\left(\dfrac{1}{2}-3\right)=
−23[45−6(2−13)]=-\dfrac{2}{3}\left[\dfrac{4}{5}-6\left(2-\dfrac{1}{3}\right)\right]=
53[13−2]215−3(26)=\dfrac{\dfrac{5}{3}\left[\dfrac{1}{3}-2\right]^2}{\dfrac{1}{5}-3\left(\dfrac{2}{6}\right)}=
Bloque 3
−83−23+74−94−315=-\dfrac{8}{3}-\dfrac{2}{3}+\dfrac{7}{4}-\dfrac{9}{4}-\dfrac{3}{15}=
23−(14)2−3(−12−34)=\dfrac{2}{3}-\left(\dfrac{1}{4}\right)^2-3\left(-\dfrac{1}{2}-\dfrac{3}{4}\right)=
−45[−23−25(34+1)]=-\dfrac{4}{5}\left[-\dfrac{2}{3}-\dfrac{2}{5}\left(\dfrac{3}{4}+1\right)\right]=
34[15−3]252−5(24)=\dfrac{\dfrac{3}{4}\left[\dfrac{1}{5}-3\right]^2}{\dfrac{5}{2}-5\left(\dfrac{2}{4}\right)}=
Bloque 4
26−35+42−35−12=\dfrac{2}{6}-\dfrac{3}{5}+\dfrac{4}{2}-\dfrac{3}{5}-\dfrac{1}{2}=
34−25(13−46)+(23)2=\dfrac{3}{4}-\dfrac{2}{5}\left(\dfrac{1}{3}-\dfrac{4}{6}\right)+\left(\dfrac{2}{3}\right)^2=
63[4−5(−32−2)]=\dfrac{6}{3}\left[4-5\left(-\dfrac{3}{2}-2\right)\right]=
−56[12−5]234−5(110)=\dfrac{-\dfrac{5}{6}\left[\dfrac{1}{2}-5\right]^2}{\dfrac{3}{4}-5\left(\dfrac{1}{10}\right)}=

Respuestas

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Bloque 1
14−65+64+35−710=74−35−710=35−12−1420=920\begin{aligned} \tfrac{1}{4}-\tfrac{6}{5}+\tfrac{6}{4}+\tfrac{3}{5}-\tfrac{7}{10} &= \\[14pt] \tfrac{7}{4}-\tfrac{3}{5}-\tfrac{7}{10} &= \\[14pt] \tfrac{35-12-14}{20} &= \mathbf{\tfrac{9}{20}} \end{aligned}
(12)3−4(32−56)−14=18−4(9−56)−14=18−166−14=3−64−624=−6724\begin{aligned} \left(\tfrac{1}{2}\right)^3-4\left(\tfrac{3}{2}-\tfrac{5}{6}\right)-\tfrac{1}{4} &= \\[14pt] \tfrac{1}{8}-4\left(\tfrac{9-5}{6}\right)-\tfrac{1}{4} &= \\[14pt] \tfrac{1}{8}-\tfrac{16}{6}-\tfrac{1}{4} &= \\[14pt] \tfrac{3-64-6}{24} &= \mathbf{-\tfrac{67}{24}} \end{aligned}
23 ⁣[12+4 ⁣(3−25)]=23 ⁣[12+4 ⁣(135)]=23 ⁣[12+525]=23 ⁣[5+10410]=10915\begin{aligned} \tfrac{2}{3}\!\left[\tfrac{1}{2}+4\!\left(3-\tfrac{2}{5}\right)\right] &= \\[14pt] \tfrac{2}{3}\!\left[\tfrac{1}{2}+4\!\left(\tfrac{13}{5}\right)\right] &= \\[14pt] \tfrac{2}{3}\!\left[\tfrac{1}{2}+\tfrac{52}{5}\right] &= \\[14pt] \tfrac{2}{3}\!\left[\tfrac{5+104}{10}\right] &= \mathbf{\tfrac{109}{15}} \end{aligned}
35[12−4]232−2 ⁣(16)=35[−72]232−26=35⋅4949−26=1472076=882140=6310\begin{aligned} \frac{\tfrac{3}{5}\left[\tfrac{1}{2}-4\right]^2}{\tfrac{3}{2}-2\!\left(\tfrac{1}{6}\right)} &= \\[14pt] \frac{\tfrac{3}{5}\left[-\tfrac{7}{2}\right]^2}{\tfrac{3}{2}-\tfrac{2}{6}} &= \\[14pt] \frac{\tfrac{3}{5}\cdot\tfrac{49}{4}}{\tfrac{9-2}{6}} &= \\[14pt] \frac{\tfrac{147}{20}}{\tfrac{7}{6}} = \tfrac{882}{140} &= \mathbf{\tfrac{63}{10}} \end{aligned}
Bloque 2
−24+63−84−73+16=−104−13+16=−30−4+212=−3212=−83\begin{aligned} -\tfrac{2}{4}+\tfrac{6}{3}-\tfrac{8}{4}-\tfrac{7}{3}+\tfrac{1}{6} &= \\[14pt] -\tfrac{10}{4}-\tfrac{1}{3}+\tfrac{1}{6} &= \\[14pt] \tfrac{-30-4+2}{12} &= \\[14pt] -\tfrac{32}{12} &= \mathbf{-\tfrac{8}{3}} \end{aligned}
(25)2−35−4 ⁣(12−3)=425−35−4 ⁣(−52)=425−35+10=8−30+50050=23925\begin{aligned} \left(\tfrac{2}{5}\right)^2-\tfrac{3}{5}-4\!\left(\tfrac{1}{2}-3\right) &= \\[14pt] \tfrac{4}{25}-\tfrac{3}{5}-4\!\left(-\tfrac{5}{2}\right) &= \\[14pt] \tfrac{4}{25}-\tfrac{3}{5}+10 &= \\[14pt] \tfrac{8-30+500}{50} &= \mathbf{\tfrac{239}{25}} \end{aligned}
−23 ⁣[45−6 ⁣(2−13)]=−23 ⁣[45−6 ⁣(53)]=−23 ⁣[45−10]=−23 ⁣[−465]=9215\begin{aligned} -\tfrac{2}{3}\!\left[\tfrac{4}{5}-6\!\left(2-\tfrac{1}{3}\right)\right] &= \\[14pt] -\tfrac{2}{3}\!\left[\tfrac{4}{5}-6\!\left(\tfrac{5}{3}\right)\right] &= \\[14pt] -\tfrac{2}{3}\!\left[\tfrac{4}{5}-10\right] &= \\[14pt] -\tfrac{2}{3}\!\left[-\tfrac{46}{5}\right] &= \mathbf{\tfrac{92}{15}} \end{aligned}
53[13−2]215−3 ⁣(26)=53[−53]215−1=53⋅259−45=12527−45=−625108\begin{aligned} \frac{\tfrac{5}{3}\left[\tfrac{1}{3}-2\right]^2}{\tfrac{1}{5}-3\!\left(\tfrac{2}{6}\right)} &= \\[14pt] \frac{\tfrac{5}{3}\left[-\tfrac{5}{3}\right]^2}{\tfrac{1}{5}-1} &= \\[14pt] \frac{\tfrac{5}{3}\cdot\tfrac{25}{9}}{-\tfrac{4}{5}} &= \\[14pt] \frac{\tfrac{125}{27}}{-\tfrac{4}{5}} &= \mathbf{-\tfrac{625}{108}} \end{aligned}
Bloque 3
−83−23+74−94−315=−103−24−315=−200−30−1260=−12130\begin{aligned} -\tfrac{8}{3}-\tfrac{2}{3}+\tfrac{7}{4}-\tfrac{9}{4}-\tfrac{3}{15} &= \\[14pt] -\tfrac{10}{3}-\tfrac{2}{4}-\tfrac{3}{15} &= \\[14pt] \tfrac{-200-30-12}{60} &= \mathbf{-\tfrac{121}{30}} \end{aligned}
23− ⁣(14)2−3 ⁣(−12−34)=23−116−3 ⁣(−54)=23−116+154=32−3+18048=20948\begin{aligned} \tfrac{2}{3}-\!\left(\tfrac{1}{4}\right)^2-3\!\left(-\tfrac{1}{2}-\tfrac{3}{4}\right) &= \\[14pt] \tfrac{2}{3}-\tfrac{1}{16}-3\!\left(-\tfrac{5}{4}\right) &= \\[14pt] \tfrac{2}{3}-\tfrac{1}{16}+\tfrac{15}{4} &= \\[14pt] \tfrac{32-3+180}{48} &= \mathbf{\tfrac{209}{48}} \end{aligned}
−45 ⁣[−23−25 ⁣(34+1)]=−45 ⁣[−23−25 ⁣(74)]=−45 ⁣[−23−1420]=−45 ⁣[−8260]=8275\begin{aligned} -\tfrac{4}{5}\!\left[-\tfrac{2}{3}-\tfrac{2}{5}\!\left(\tfrac{3}{4}+1\right)\right] &= \\[14pt] -\tfrac{4}{5}\!\left[-\tfrac{2}{3}-\tfrac{2}{5}\!\left(\tfrac{7}{4}\right)\right] &= \\[14pt] -\tfrac{4}{5}\!\left[-\tfrac{2}{3}-\tfrac{14}{20}\right] &= \\[14pt] -\tfrac{4}{5}\!\left[-\tfrac{82}{60}\right] &= \mathbf{\tfrac{82}{75}} \end{aligned}
34[15−3]252−5 ⁣(24)=34[−145]252−52=34⋅196250=Indeterminado\begin{aligned} \frac{\tfrac{3}{4}\left[\tfrac{1}{5}-3\right]^2}{\tfrac{5}{2}-5\!\left(\tfrac{2}{4}\right)} &= \\[14pt] \frac{\tfrac{3}{4}\left[-\tfrac{14}{5}\right]^2}{\tfrac{5}{2}-\tfrac{5}{2}} &= \\[14pt] \frac{\tfrac{3}{4}\cdot\tfrac{196}{25}}{0} &= \mathbf{\text{Indeterminado}} \end{aligned}
Bloque 4
26−35+42−35−12=13−65+32=10−36+4530=1930\begin{aligned} \tfrac{2}{6}-\tfrac{3}{5}+\tfrac{4}{2}-\tfrac{3}{5}-\tfrac{1}{2} &= \\[14pt] \tfrac{1}{3}-\tfrac{6}{5}+\tfrac{3}{2} &= \\[14pt] \tfrac{10-36+45}{30} &= \mathbf{\tfrac{19}{30}} \end{aligned}
34−25 ⁣(13−46)+ ⁣(23)2=34−25 ⁣(−13)+49=34+215+49=135+24+80180=239180\begin{aligned} \tfrac{3}{4}-\tfrac{2}{5}\!\left(\tfrac{1}{3}-\tfrac{4}{6}\right)+\!\left(\tfrac{2}{3}\right)^2 &= \\[14pt] \tfrac{3}{4}-\tfrac{2}{5}\!\left(-\tfrac{1}{3}\right)+\tfrac{4}{9} &= \\[14pt] \tfrac{3}{4}+\tfrac{2}{15}+\tfrac{4}{9} &= \\[14pt] \tfrac{135+24+80}{180} &= \mathbf{\tfrac{239}{180}} \end{aligned}
63 ⁣[4−5 ⁣(−32−2)]=2 ⁣[4−5 ⁣(−72)]=2 ⁣[4+352]=2 ⁣[432]=43\begin{aligned} \tfrac{6}{3}\!\left[4-5\!\left(-\tfrac{3}{2}-2\right)\right] &= \\[14pt] 2\!\left[4-5\!\left(-\tfrac{7}{2}\right)\right] &= \\[14pt] 2\!\left[4+\tfrac{35}{2}\right] &= \\[14pt] 2\!\left[\tfrac{43}{2}\right] &= \mathbf{43} \end{aligned}
−56[12−5]234−5 ⁣(110)=−56[−92]234−510=−56⋅81414=−4052414=−162024=−1352\begin{aligned} \frac{-\tfrac{5}{6}\left[\tfrac{1}{2}-5\right]^2}{\tfrac{3}{4}-5\!\left(\tfrac{1}{10}\right)} &= \\[14pt] \frac{-\tfrac{5}{6}\left[-\tfrac{9}{2}\right]^2}{\tfrac{3}{4}-\tfrac{5}{10}} &= \\[14pt] \frac{-\tfrac{5}{6}\cdot\tfrac{81}{4}}{\tfrac{1}{4}} &= \\[14pt] \frac{-\tfrac{405}{24}}{\tfrac{1}{4}} = -\tfrac{1620}{24} &= \mathbf{-\tfrac{135}{2}} \end{aligned}

Ejercicios extra

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Bloque A
73+34+65−84−25=12760\dfrac{7}{3} + \dfrac{3}{4} + \dfrac{6}{5} - \dfrac{8}{4} - \dfrac{2}{5} = \mathbf{\dfrac{127}{60}}
−35+(25)2−4 ⁣(−13−45)=30775-\dfrac{3}{5} + \left(\dfrac{2}{5}\right)^2 - 4\!\left(-\dfrac{1}{3} - \dfrac{4}{5}\right) = \mathbf{\dfrac{307}{75}}
56 ⁣[−34+36 ⁣(45−1)]=−1724\dfrac{5}{6}\!\left[-\dfrac{3}{4} + \dfrac{3}{6}\!\left(\dfrac{4}{5} - 1\right)\right] = \mathbf{-\dfrac{17}{24}}
−45+32+45+38−68=98-\dfrac{4}{5} + \dfrac{3}{2} + \dfrac{4}{5} + \dfrac{3}{8} - \dfrac{6}{8} = \mathbf{\dfrac{9}{8}}
−64+45−56−74−112=−103-\dfrac{6}{4} + \dfrac{4}{5} - \dfrac{5}{6} - \dfrac{7}{4} - \dfrac{1}{12} = \mathbf{-\dfrac{10}{3}}
47+(−36)2−5 ⁣(−24+56)=−2328\dfrac{4}{7} + \left(-\dfrac{3}{6}\right)^2 - 5\!\left(-\dfrac{2}{4} + \dfrac{5}{6}\right) = \mathbf{-\dfrac{23}{28}}
−67 ⁣[−45−47 ⁣(56+1)]=65-\dfrac{6}{7}\!\left[-\dfrac{4}{5} - \dfrac{4}{7}\!\left(\dfrac{5}{6} + 1\right)\right] = \mathbf{\dfrac{6}{5}}
58−(47)2+6 ⁣(−35+67)=717392\dfrac{5}{8} - \left(\dfrac{4}{7}\right)^2 + 6\!\left(-\dfrac{3}{5} + \dfrac{6}{7}\right) = \mathbf{\dfrac{717}{392}}
Bloque B
12+14−18=58\dfrac{1}{2} + \dfrac{1}{4} - \dfrac{1}{8} = \mathbf{\dfrac{5}{8}}
35÷23+110=1\dfrac{3}{5} \div \dfrac{2}{3} + \dfrac{1}{10} = \mathbf{1}
(23−14)⋅125=1\left(\dfrac{2}{3} - \dfrac{1}{4}\right) \cdot \dfrac{12}{5} = \mathbf{1}
43−[12⋅(14+13)]=2524\dfrac{4}{3} - \left[\dfrac{1}{2}\cdot\left(\dfrac{1}{4} + \dfrac{1}{3}\right)\right] = \mathbf{\dfrac{25}{24}}
34−1218+14=23\dfrac{\dfrac{3}{4} - \dfrac{1}{2}}{\dfrac{1}{8} + \dfrac{1}{4}} = \mathbf{\dfrac{2}{3}}
(12)2+34⋅13=12\left(\dfrac{1}{2}\right)^2 + \dfrac{3}{4}\cdot\dfrac{1}{3} = \mathbf{\dfrac{1}{2}}
25⋅(12+13)−110=730\dfrac{2}{5}\cdot\left(\dfrac{1}{2} + \dfrac{1}{3}\right) - \dfrac{1}{10} = \mathbf{\dfrac{7}{30}}
56÷(1−13)+14=32\dfrac{5}{6} \div \left(1 - \dfrac{1}{3}\right) + \dfrac{1}{4} = \mathbf{\dfrac{3}{2}}
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