\(\dfrac{333}{777}\) + \(\dfrac{22}{55}\)
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\(\dfrac{11}{6}+\dfrac{1}{4}=\dfrac{22}{12}+\dfrac{3}{12}=\dfrac{25}{12}\)
\(\dfrac{2}{5}-\dfrac{3}{8}=\dfrac{16}{40}-\dfrac{15}{40}=\dfrac{1}{40}\)
\(\dfrac{3}{10}-\dfrac{4}{15}=\dfrac{9}{30}-\dfrac{8}{30}=\dfrac{1}{30}\)
\(3+\dfrac{2}{5}=\dfrac{15}{5}+\dfrac{2}{5}=\dfrac{17}{5}\)
\(\dfrac{333}{777}+\dfrac{22}{55}=\dfrac{3}{7}+\dfrac{2}{5}=\dfrac{15}{35}+\dfrac{14}{35}=\dfrac{29}{35}\)
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a) Ta có: 333777 = 333111.7 = (7773)111
777333 = 777111.3 = (7773)111
Vì 7773<3337 nên (7773)111 < (7773)111
Vậy 333777 > 777333
b) Ta có: 2222 = 22.111 =(2111)2
2222 = 2211.2 = (2211)2
Vì 2111 > 2211 nên (2111)2 > (2211)2
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= 2/5 + 1/5 + 1/5
= 4/5
b. = 2/7 + 4/7 + 5/21
= 6/21 + 12/21 + 5/21
= 23/21
HT
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a) \(\dfrac{22}{55}=\dfrac{2}{5}\)
b) \(\dfrac{-63}{81}=\dfrac{-7}{9}\)
c) \(\dfrac{2.14}{7.8}=\dfrac{2.7.2}{7.2.2.2}=\dfrac{1}{2}\)
d) \(\dfrac{49+7.49}{49}=\dfrac{49.\left(7+1\right)}{49}=\dfrac{49.8}{49}=8\)
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a)Ta có:
\(\dfrac{-3}{7}=\dfrac{\left(-3\right)\cdot111}{7.111}=\dfrac{-333}{7}\)
b)\(15\%=\dfrac{3}{20}=0,15\)
\(5\%=\dfrac{1}{20}=0,05\)
c)\(-5\dfrac{1}{2}=-\dfrac{11}{2}\Rightarrow\)số nghịch đảo của nó là:\(-\dfrac{2}{11}\)
\(\)\(1,3=\dfrac{13}{10}\Rightarrow\)số nghịch đảo của nó là:\(\dfrac{10}{13}\)
\(\dfrac{333}{777}+\dfrac{22}{55}=\dfrac{333:111}{777:111}+\dfrac{22:11}{55:11}\\ =\dfrac{3}{7}+\dfrac{2}{5}=\dfrac{3\times5+2\times7}{35}\\ =\dfrac{15+14}{35}=\dfrac{29}{35}\)