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a) \(\dfrac{4.7}{9.32}=\dfrac{4.7}{9.4.8}=\dfrac{7}{72}\) b) \(\dfrac{3.21}{14.15}=\dfrac{3.7.3}{7.2.3.5}=\dfrac{3}{10}\)
c) \(\dfrac{2.5.13}{26.35}=\dfrac{2.5.13}{2.13.5.7}=\dfrac{1}{7}\) d)\(\dfrac{9.6-9.3}{18}\) \(=\dfrac{9.\left(6-3\right)}{9.2}=\dfrac{9.3}{9.2}=\dfrac{3}{2}\)
e) \(\dfrac{17.5-17}{3-20}=\dfrac{17.\left(5-1\right)}{-17}=-\dfrac{17.4}{17}=-4\) f) \(\dfrac{49+7.49}{49}=\dfrac{49.\left(1+7\right)}{49}=8\)
Bài 1 : Rút gọn các phân số sau đến tối giản :
a) \(\dfrac{3.21}{14.15}=\dfrac{3.3.7}{2.7.3.5}=\dfrac{1.3.1}{2.1.1.5}=\dfrac{3}{10}\)
b) \(\dfrac{49+49.7}{49}=\dfrac{49\left(1+7\right)}{49}=\dfrac{49.8}{49}=\dfrac{1.8}{1}=\dfrac{8}{1}=8\)
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7)\(\dfrac{-19}{34}\left(\dfrac{17}{19}+\dfrac{49}{18}\right)+\dfrac{49}{18}\left(\dfrac{19}{34}-\dfrac{18}{7}\right)\)
=\(\dfrac{-19}{34}.\dfrac{17}{19}+\dfrac{49}{18}.\dfrac{-19}{34}+\dfrac{49}{18}.\dfrac{19}{34}-\dfrac{18}{7}.\dfrac{49}{18}\)
=\(\dfrac{1}{2}+\left(\dfrac{49}{18}.\dfrac{-19}{34}+\dfrac{49}{18}.\dfrac{19}{34}\right)-7\)
=\(\dfrac{1}{2}+\left[\dfrac{49}{18}\left(\dfrac{-19}{34}+\dfrac{19}{34}\right)\right]-7\)
=\(\dfrac{1}{2}+0-7=\dfrac{-13}{2}\)
8)\(\dfrac{29}{32}\left(\dfrac{41}{36}-\dfrac{32}{58}\right)-\dfrac{41}{36}\left(\dfrac{29}{32}+\dfrac{18}{41}\right)\)
=\(\dfrac{29}{32}.\dfrac{41}{36}-\dfrac{29}{32}.\dfrac{32}{58}-\dfrac{41}{36}.\dfrac{29}{32}+\dfrac{18}{41}.\dfrac{41}{36}\)
=\(\left(\dfrac{29}{32}.\dfrac{41}{36}-\dfrac{41}{36}\dfrac{29}{32}\right)-\dfrac{29}{32}.\dfrac{32}{58}+\dfrac{18}{41}.\dfrac{41}{36}\)
=\(0-\dfrac{1}{2}+\dfrac{1}{2}=0\)
Tính các tổng dưới đây sau khi đã rút gọn phân số :
a)\(\dfrac{7}{21}\) + \(\dfrac{9}{-36}\) = \(\dfrac{7}{21}\)+\(\dfrac{-9}{36}\)=\(\dfrac{1}{3}\)+\(\dfrac{-1}{4}\)=\(\dfrac{4}{12}\)+\(\dfrac{-3}{12}\)=\(\dfrac{1}{12}\)
b) \(\dfrac{-12}{18}\)+\(\dfrac{-21}{35}\)=\(\dfrac{-2}{3}\)+\(\dfrac{-3}{5}\)=\(\dfrac{-10}{15}\)+\(\dfrac{-9}{15}\)=\(\dfrac{-19}{15}\)
c) \(\dfrac{-3}{21}\)+\(\dfrac{6}{42}\)=\(\dfrac{-1}{7}\)+\(\dfrac{1}{7}\)=0
d) \(\dfrac{-18}{24}\)+\(\dfrac{15}{-21}\)=\(\dfrac{-18}{24}\)+\(\dfrac{-15}{21}\)=\(\dfrac{-3}{4}\)+\(\dfrac{-5}{7}\)=\(\dfrac{-21}{28}\)+\(\dfrac{-20}{28}\)=\(\dfrac{-41}{28}\)
a)\(12< 13;49>47\)
\(\Rightarrow\dfrac{12}{49}< \dfrac{13}{47}\)
b)\(\dfrac{64}{85}>\dfrac{43}{85}\Rightarrow\dfrac{64}{85}>\dfrac{1}{2}\)
\(\dfrac{17}{35}< \dfrac{17}{34}\Rightarrow\dfrac{17}{35}< \dfrac{1}{2}\)
\(\Rightarrow\dfrac{17}{35}< \dfrac{64}{85}\)
c) \(\dfrac{19}{31}>\dfrac{16}{31}\Rightarrow\dfrac{19}{31}>\dfrac{1}{2}\)
\(\dfrac{17}{35}< \dfrac{17}{34}\Rightarrow\dfrac{17}{35}< \dfrac{1}{2}\)
\(\Rightarrow\dfrac{17}{35}< \dfrac{19}{31}\)
d)
\(1-\dfrac{67}{77}=\dfrac{10}{77}\)
\(1-\dfrac{73}{83}=\dfrac{10}{83}\)
\(\dfrac{10}{77}>\dfrac{10}{83}\Rightarrow\dfrac{67}{77}< \dfrac{73}{83}\)
e)\(1-\dfrac{456}{461}=\dfrac{5}{461}\)
\(1-\dfrac{123}{128}=\dfrac{5}{128}\)
\(\dfrac{5}{461}< \dfrac{5}{128}\Rightarrow\dfrac{456}{461}>\dfrac{123}{128}\)
\(a,\dfrac{12}{49}< \dfrac{12}{47}< \dfrac{13}{47}\Rightarrow\dfrac{12}{49}< \dfrac{12}{47}\)
b, Ta có: \(\dfrac{17}{35}=\dfrac{51}{105}\)
\(\dfrac{64}{85}>\dfrac{64}{105}>\dfrac{51}{105}\Rightarrow\dfrac{64}{85}>\dfrac{51}{105}\) hay \(\dfrac{64}{85}>\dfrac{17}{85}\)
c,\(\dfrac{19}{31}>\dfrac{17}{31}>\dfrac{17}{35}\Rightarrow\dfrac{19}{31}>\dfrac{17}{35}\)
d, \(\dfrac{67}{77}+\dfrac{10}{77}=1\)
\(\dfrac{73}{83}+\dfrac{10}{83}=1\)
\(\dfrac{10}{77}>\dfrac{10}{83}\Rightarrow\dfrac{67}{77}< \dfrac{73}{83}\)
e, \(\dfrac{456}{461}+\dfrac{5}{461}=1\)
\(\dfrac{123}{128}+\dfrac{5}{128}=1\)
\(\dfrac{5}{461}< \dfrac{5}{128}\Rightarrow\dfrac{456}{461}>\dfrac{123}{128}\)
a) \(\dfrac{3}{4}+\dfrac{3}{5}-\dfrac{18}{60}\) ( MTC: 60)
= \(\dfrac{3.15}{4.15}+\dfrac{3.12}{5.12}-\dfrac{18}{60}\)
= \(\dfrac{45}{60}+\dfrac{36}{60}-\dfrac{18}{60}\)
= \(\dfrac{45+36-18}{60}\)=\(\dfrac{63}{60}=\dfrac{21}{20}\)
b)\(\dfrac{17}{8}-\dfrac{11}{6}-\dfrac{2}{9}\) (MTC:72)
=\(\dfrac{17.9}{8.9}-\dfrac{11.12}{6.12}-\dfrac{2.8}{9.8}\)
= \(\dfrac{153}{72}-\dfrac{132}{72}-\dfrac{16}{72}\)
=\(\dfrac{153-132-16}{72}\)
=\(\dfrac{5}{72}\)
c)\(\dfrac{23}{29}+\dfrac{5}{11}+\dfrac{17}{11}\) (MTC:319)
= \(\dfrac{23.11}{29.11}+\dfrac{5.29}{11.29}+\dfrac{17.29}{11.29}\)
=\(\dfrac{253}{319}+\dfrac{145}{319}+\dfrac{493}{319}\)
=\(\dfrac{253+145+493}{319}\)=\(\dfrac{891}{319}=\dfrac{81}{29}\)
c) \(\dfrac{20}{45}+\dfrac{14}{35}+\dfrac{32}{44}\)
= \(\dfrac{4}{9}+\dfrac{2}{5}+\dfrac{8}{11}\)(Rút gọn b/thức)(MTC:495)
=\(\dfrac{4.55}{9.55}+\dfrac{2.99}{5.99}+\dfrac{8.45}{11.45}\)
=\(\dfrac{220}{495}+\dfrac{198}{495}+\dfrac{360}{495}\)
=\(\dfrac{220+198+360}{495}\)=\(\dfrac{778}{495}\)
e)\(17\dfrac{25}{27}+3\dfrac{7}{2}\)
= \(\dfrac{484}{27}+\dfrac{13}{2}\) (MTC:54)
=\(\dfrac{484.2}{27.2}+\dfrac{13.27}{2.27}\)
\(=\dfrac{968}{54}+\dfrac{351}{54}\)
=\(\dfrac{968+351}{54}=\dfrac{1319}{54}\)
a) \(\dfrac{32}{12}=\dfrac{4.8}{4.3}=\dfrac{8}{3}\)
b) \(\dfrac{-26}{156}=\dfrac{-26}{26.6}=\dfrac{-1}{6}\)
c) \(\dfrac{4.7}{9.32}=\dfrac{4.7}{9.4.8}=\dfrac{7}{9.8}=\dfrac{7}{72}\)
d) \(\dfrac{2.5.13}{26.35}\dfrac{2.5.13}{2.13.5.7}=\dfrac{1}{7}\)
e)\(\dfrac{7.6-9.13}{18}=\dfrac{42-117}{18}=\dfrac{-75}{18}=\dfrac{-25.3}{6.3}=\dfrac{-25}{6}\)
f) \(\dfrac{17.5-17}{3-20}=\dfrac{17.\left(5-1\right)}{-17}=-4\)
g) \(\dfrac{49+7.49}{49}=\dfrac{49.\left(1+7\right)}{49}=8\)
a, \(\dfrac{32}{12}=\dfrac{8}{3}\)
b, \(\dfrac{-26}{156}=\dfrac{-1}{6}\)
c, \(\dfrac{4.7}{9.32}=\dfrac{4.7}{9.4.8}=\dfrac{7}{72}\)
d,\(\dfrac{2.5.13}{26.35}=\dfrac{2.5.13}{2.13.7.5}=\dfrac{1}{7}\)
e,\(\dfrac{7.6-9.13}{18}=\dfrac{3\left(7.2-3.13\right)}{18}=\dfrac{-25}{6}\)
f ,\(\dfrac{17.5-17}{3-20}=\dfrac{17\left(5-1\right)}{-17}=-4\)
g,\(\dfrac{49+7.49}{49}=\dfrac{49\left(1+7\right)}{49}=8\)