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a , \(\left(\dfrac{-2}{3}+1\dfrac{1}{4}-\dfrac{1}{6}\right):\dfrac{-24}{10}\)
=\(\left(\dfrac{-2}{3}+\dfrac{5}{4}-\dfrac{1}{6}\right):\dfrac{-12}{5}\)
=\(\left(\dfrac{-8}{12}+\dfrac{15}{12}-\dfrac{2}{12}\right)\cdot\dfrac{-5}{12}\)
=\(\dfrac{5}{12}\cdot\dfrac{-5}{12}=\dfrac{-25}{144}\)
b , \(\dfrac{13}{15}\cdot0,25\cdot3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right)1\dfrac{23}{24}\)
=\(\dfrac{13}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right)\cdot\dfrac{57}{24}\)
=\(\dfrac{13}{20}-\dfrac{47}{60}\cdot\dfrac{57}{24}\)
=\(\dfrac{13}{20}-\dfrac{893}{480}=\dfrac{312}{480}-\dfrac{893}{480}=\dfrac{-581}{480}\)
c , \(\left(\dfrac{12}{32}+\dfrac{5}{-20}-\dfrac{10}{24}\right):\dfrac{2}{3}\)
=\(\left(\dfrac{180}{480}-\dfrac{120}{480}-\dfrac{200}{480}\right)\cdot\dfrac{3}{2}\)
= \(\dfrac{-7}{24}\cdot\dfrac{3}{2}=\dfrac{-7}{16}\)
d , \(4\dfrac{1}{2}:\left(2,5-3\dfrac{3}{4}\right)+\left(-\dfrac{1}{2}\right)\)
=\(\dfrac{9}{2}:\left(\dfrac{5}{2}-\dfrac{15}{4}\right)-\dfrac{1}{2}\)
=\(\dfrac{9}{2}:\dfrac{-5}{4}-\dfrac{1}{2}=\dfrac{9}{2}\cdot\dfrac{-4}{5}-\dfrac{1}{2}=\dfrac{-18}{5}-\dfrac{1}{2}=\dfrac{-41}{10}\)
e , \(\dfrac{-5}{2}:\left(\dfrac{3}{4}-\dfrac{1}{2}\right)=\dfrac{-5}{2}\left(\dfrac{3}{4}-\dfrac{2}{4}\right)\)
=\(\dfrac{-5}{2}:\dfrac{1}{4}=\dfrac{-5}{2}\cdot4=-10\)
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)
=\(\dfrac{484.2}{27.2}+\dfrac{13.27}{2.27}\)
\(=\dfrac{968}{54}+\dfrac{351}{54}\)
=\(\dfrac{968+351}{54}=\dfrac{1319}{54}\)
2) Tinh nhanh:
a) \(\dfrac{5}{23}\) . \(\dfrac{17}{26}\) + \(\dfrac{5}{23}\) . \(\dfrac{10}{26}\) - \(\dfrac{5}{23}\)
= \(\dfrac{5}{23}\) . \(\left(\dfrac{17}{26}+\dfrac{10}{26}-1\right)\)
= \(\dfrac{5}{23}\) . \(\left(\dfrac{27}{26}-1\right)\) = \(\dfrac{5}{23}\) . \(\dfrac{1}{26}\)
= \(\dfrac{5}{598}\)
b) \(\dfrac{1}{7}.\dfrac{5}{9}+\dfrac{5}{9}.\dfrac{2}{7}+\dfrac{5}{9}.\dfrac{1}{7}+\dfrac{5}{9}.\dfrac{3}{7}\)
= \(\dfrac{5}{9}.\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)
= \(\dfrac{5}{9}\) . 1= \(\dfrac{5}{9}\)
Ta có: \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{19}+\dfrac{1}{20}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{20}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{19}+\dfrac{1}{20}-\left(1+\dfrac{1}{2}+...+\dfrac{1}{10}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{19}+\dfrac{1}{20}-1-\dfrac{1}{2}-...-\dfrac{1}{10}\)
\(=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\)
Vậy \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\)
1:-5/14=-50/140
3/20=21/140
9/70=18/140
2: -55/132=-5/12=-35/84
10/42=20/84
-3/28=-9/84
3: 7/10=231/330
1/33=10/330
E=\(\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\\ E=\dfrac{1}{90}-\left(\dfrac{1}{72}+\dfrac{1}{56}+\dfrac{1}{42}+\dfrac{1}{30}+\dfrac{1}{20}+\dfrac{1}{12}+\dfrac{1}{6}+\dfrac{1}{2}\right)\\ E=\dfrac{1}{90}-\left(\dfrac{1}{9.8}+\dfrac{1}{8.7}+\dfrac{1}{7.6}+\dfrac{1}{6.5}+\dfrac{1}{5.4}+\dfrac{1}{4.3}+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\\ E=\dfrac{1}{90}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)\\ E=\dfrac{1}{90}-\left(1-\dfrac{1}{9}\right)\\ E=\dfrac{1}{90}-\dfrac{8}{9}\\ E=\dfrac{1}{90}-\dfrac{80}{90}\\ E=-\dfrac{79}{90}\)Vậy:\(E=-\dfrac{79}{90}\)
E=\(\dfrac{1}{10.9}-\dfrac{1}{9.8}-\dfrac{1}{8.7}-\dfrac{1}{7.6}-\dfrac{1}{6.5}-\dfrac{1}{5.4}-\dfrac{1}{4.3}-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
E=\(\dfrac{1}{10}-\dfrac{1}{1}\)
E=\(\dfrac{-9}{10}\)
Giải:
a) \(A=\dfrac{5}{13}.\dfrac{5}{7}+\dfrac{-20}{41}+\dfrac{5}{13}+\dfrac{-21}{41}\)
\(\Leftrightarrow A=\dfrac{5}{13}.\dfrac{5}{7}+\dfrac{5}{13}+\dfrac{-21}{41}+\dfrac{-20}{41}\)
\(\Leftrightarrow A=\dfrac{5}{13}\left(\dfrac{5}{7}+1\right)+\dfrac{-41}{41}\)
\(\Leftrightarrow A=\dfrac{5}{13}.\dfrac{12}{7}+\left(-1\right)\)
\(\Leftrightarrow A=\dfrac{60}{91}+\left(-1\right)=-\dfrac{31}{91}\)
Vậy ...
b) \(B=\dfrac{5}{7}.\dfrac{2}{11}+\dfrac{5}{7}.\dfrac{12}{11}-\dfrac{5}{7}.\dfrac{7}{11}\)
\(\Leftrightarrow B=\dfrac{5}{7}\left(\dfrac{2}{11}+\dfrac{12}{11}-\dfrac{7}{11}\right)\)
\(\Leftrightarrow B=\dfrac{5}{7}.\dfrac{7}{11}\)
\(\Leftrightarrow B=\dfrac{5}{11}\)
Vậy ...
c) \(C=\dfrac{-2}{3}+\dfrac{-5}{7}+\dfrac{2}{3}+\dfrac{-2}{7}\)
\(\Leftrightarrow C=\left(\dfrac{-2}{3}+\dfrac{2}{3}\right)+\left(\dfrac{-2}{7}+\dfrac{-5}{7}\right)\)
\(\Leftrightarrow C=0+\left(-1\right)=-1\)
Vậy ...
Ta có:\(\dfrac{31}{2}\).\(\dfrac{32}{2}\).\(\dfrac{33}{2}\).....\(\dfrac{60}{2}\)
=\(\dfrac{31.32.33.....60}{2^{30}}\)
=\(\dfrac{\left(1.2.3.....30\right).\left(31.32.33.....60\right)}{\left(1.2.3.....30\right).2^{30}}\)
=\(\dfrac{1.2.3.....60}{2.4.6.....60}\)
=\(\dfrac{\left(1.3.5.....59\right).\left(2.4.6.....60\right)}{2.4.6.....60}\)
=1.3.5.....59
Vậy (đpcm)