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\(\frac{125}{34}+\frac{78}{12}=10\frac{3}{17}=\frac{173}{17}\)
\(\frac{12}{78}\times\frac{23}{12}=\frac{23}{78}\)
\(\frac{98}{24}-\frac{12}{97}=3\frac{1117}{1164}=\frac{4609}{1164}\)
\(\frac{12}{33}:\frac{5}{12}=\frac{48}{55}\)
-----------HOK TỐT----------
Thứ tự từ bé đến lớn:
\(\frac{12}{12}\)( =1) ; \(\frac{12}{6}\)( =2) ; \(\frac{12}{3}\)( =4) ; \(\frac{12}{2}\)( =6)
\(\frac{12}{12}\)\(\frac{12}{6}\)\(\frac{12}{3}\)\(\frac{12}{2}\)
\(\frac{12}{20}\)= \(\frac{6}{10}\)= \(\frac{3}{5}\)
\(\frac{3}{4}\)= \(\frac{9}{12}\)= \(\frac{12}{16}\)= \(\frac{15}{20}\)
\(\frac{75}{45}\)= \(\frac{15}{9}\)= \(\frac{5}{3}\)
\(\frac{3}{4}\)= \(\frac{18}{24}\)= \(\frac{12}{16}\)
12/20 = 6/10 = 3/5
3/4 = 9/12 = 12/16 = 15/20
75/45 = 15/9 = 5/3
3/4 = 18/24 = 12/16
Tk mk nha
\(\frac{3}{12}+\frac{2}{4}-\frac{2}{12}\)
\(=\frac{3}{12}+\frac{6}{12}-\frac{2}{12}\)
\(=\frac{3+6-2}{12}\)
\(=\frac{7}{12}\)
\(\frac{3}{12}+\frac{2}{4}-\frac{2}{12}\)
\(=\frac{3}{12}+\frac{6}{12}-\frac{2}{12}\)
\(=\frac{9}{12}-\frac{2}{12}\)
\(=\frac{7}{12}\)
54/12:64/12=27/32
56/45x54/51=112/85
56/20+58/21=584/105
21/78-12/78=3/26
\(7:\frac{1}{5}=7\cdot\frac{5}{1}=35\)
\(\frac{5}{12}-\frac{5}{15}=\frac{5\cdot15-5\cdot12}{12\cdot15}=\frac{1}{12}\)
\(\frac{7}{9}:\frac{5}{12}=\frac{7}{9}\cdot\frac{12}{5}=\frac{28}{15}\)
\(7:\frac{1}{5}=35\)
\(\frac{5}{12}-\frac{5}{15}=\frac{1}{12}\)
\(\frac{7}{9}:\frac{5}{12}=1\frac{13}{15}=\frac{28}{15}\)
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
lớp 4 mà học dạng này rồi sao!?
\(\frac{12}{3\cdot7}+\frac{12}{7\cdot11}+...+\frac{12}{43\cdot47}=3\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{4}{43\cdot47}\right)=3\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{43}-\frac{1}{47}\right)\)\(=3\left[\left(\frac{1}{3}-\frac{1}{47}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{43}-\frac{1}{43}\right)\right]=3\left[\left(\frac{47}{141}-\frac{3}{141}\right)+0+...+0\right]=3\cdot\frac{44}{141}=\frac{44}{47}\)