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a,1+1+2+2+3+3+...+100+100
=1x2+2x2+3x2+...+100x2
=2x(1+2+3+...+100)
=\(2.\frac{\left(100+1\right).\left[\left(100-1\right):1+1\right]}{2}\)
=2x5050
=10100
Chú ý dấu . là x
a, 1+1+2+2+3+...+100+100
=1+2+3+...+100+1+2+3+...+100
= (100+1)*50 /2 + (100+1)*50/2
=5050+5050
=11000
\(\frac{2}{5}=\frac{12}{30}=\frac{40}{100}\)
\(\frac{4}{7}=\frac{12}{21}=\frac{20}{35}\)
nha
anh bn
\(\frac{2}{5}\)\(=\)\(\frac{12}{30}\)\(;\)\(\frac{4}{7}\)\(=\)\(\frac{12}{21}\)\(Done\)
Mình sửa lại đề xíu.
a) \(\frac{75}{100}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}=\frac{3}{4}+\frac{1}{4}+\frac{18}{21}+\frac{3}{21}+\frac{19}{32}+\frac{13}{32}=1+1+1=3\)
b) \(4\frac{2}{5}+5\frac{6}{9}+2\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}=4+\frac{2}{5}+\frac{3}{5}+5+\frac{2}{3}+\frac{1}{3}+2+\frac{3}{4}+\frac{1}{4}\)
\(=4+1+5+1+2+1=14.\)
c) \(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{41\cdot43}=\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+\frac{9-7}{7\cdot9}+...+\frac{43-41}{41\cdot43}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{41}-\frac{1}{43}=\frac{1}{3}-\frac{1}{43}=\frac{43-3}{3\cdot43}=\frac{40}{129}.\)
- a B,b D
- a \(\frac{1}{2}\)b \(\frac{2}{5}\)
- \(\frac{2}{3};\frac{10}{17};\frac{5}{11};\frac{4}{9}\)
- a\(\frac{5}{12}\)b\(\frac{97}{36}\)
a) \(\frac{4}{7}=\frac{16}{28}\)
\(\frac{9}{12}=\frac{3}{4}=\frac{21}{28}\)
b) \(\frac{13}{12}=\frac{39}{36}\)
\(\frac{19}{18}=\frac{38}{36}\)
c) \(\frac{1}{5}=\frac{2}{10}\)
d) \(\frac{1}{3}=\frac{21}{63}\)
\(\frac{2}{7}=\frac{18}{63}\)
\(\frac{4}{9}=\frac{28}{63}\)
a/ 4/7 = 1-3/7 và 9/12 = 1-3/12
vì 3/7>3/12 nên 1-3/7<1-3/12
Vậy 4/7<9/12
b/ 13/12 = 1+1/12 và 19/18 = 1+1/18
Vì 1/12>1/18 nên 13/12>19/18
1.A
2.C
HT
A.\(\frac{101}{100}\)
C. \(\frac{1}{2}\)