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\(\frac{3}{12}+\frac{2}{4}-\frac{2}{12}\)
\(=\frac{3}{12}+\frac{6}{12}-\frac{2}{12}\)
\(=\frac{3+6-2}{12}\)
\(=\frac{7}{12}\)
\(\frac{3}{12}+\frac{2}{4}-\frac{2}{12}\)
\(=\frac{3}{12}+\frac{6}{12}-\frac{2}{12}\)
\(=\frac{9}{12}-\frac{2}{12}\)
\(=\frac{7}{12}\)
\(\frac{125}{34}+\frac{78}{12}=10\frac{3}{17}=\frac{173}{17}\)
\(\frac{12}{78}\times\frac{23}{12}=\frac{23}{78}\)
\(\frac{98}{24}-\frac{12}{97}=3\frac{1117}{1164}=\frac{4609}{1164}\)
\(\frac{12}{33}:\frac{5}{12}=\frac{48}{55}\)
-----------HOK TỐT----------
=( 1/12+9/12+2/12+1/12) + ( 2/7+5/7) + (10/20+9/20)
= 13/12 +1+19/20
đến đây bạn tự tính nha
\(\frac{2}{5}=\frac{12}{30}=\frac{40}{100}\)
\(\frac{4}{7}=\frac{12}{21}=\frac{20}{35}\)
nha
anh bn
\(\frac{2}{5}\)\(=\)\(\frac{12}{30}\)\(;\)\(\frac{4}{7}\)\(=\)\(\frac{12}{21}\)\(Done\)
Thứ tự từ bé đến lớn:
\(\frac{12}{12}\)( =1) ; \(\frac{12}{6}\)( =2) ; \(\frac{12}{3}\)( =4) ; \(\frac{12}{2}\)( =6)
\(\frac{12}{12}\)\(\frac{12}{6}\)\(\frac{12}{3}\)\(\frac{12}{2}\)
a) \(\frac{12}{25}+\frac{3}{5}+\frac{13}{25}\)
= \(\left(\frac{12}{25}+\frac{13}{25}\right)+\frac{3}{5}\)
= \(\frac{25}{25}+\frac{3}{5}\)
= \(1+\frac{3}{5}\)
= \(\frac{1}{1}+\frac{3}{5}\)
= \(\frac{5}{5}+\frac{3}{5}\)
= \(\frac{8}{5}\)
bài này tớ mà làm sai thì tớ ......................ko lm người........-_-
a) \(\frac{12}{25}+\frac{3}{5}+\frac{13}{25}\)
= \(\frac{12}{25}+\frac{15}{25}+\frac{13}{25}\)
= \(\frac{12+15+13}{25}\)
= \(\frac{40}{25}\)
= \(\frac{8}{5}\)
b) \(\frac{3}{2}+\frac{2}{3}+\frac{4}{3}\)
= \(\frac{9}{6}+\frac{4}{6}+\frac{8}{6}\)
= \(\frac{9+4+8}{6}\)
= \(\frac{21}{6}\)
= \(\frac{7}{2}\)
\(\left(\frac{5}{12}+\frac{9}{12}\right)+\frac{5}{12}=\frac{14}{12}+\frac{5}{12}\)
\(=\frac{19}{12}\)
lớp 4 mà học dạng này rồi sao!?
\(\frac{12}{3\cdot7}+\frac{12}{7\cdot11}+...+\frac{12}{43\cdot47}=3\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{4}{43\cdot47}\right)=3\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{43}-\frac{1}{47}\right)\)\(=3\left[\left(\frac{1}{3}-\frac{1}{47}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{43}-\frac{1}{43}\right)\right]=3\left[\left(\frac{47}{141}-\frac{3}{141}\right)+0+...+0\right]=3\cdot\frac{44}{141}=\frac{44}{47}\)