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a) $\frac{2}{3} - \frac{1}{3} = \frac{{2 - 1}}{3} = \frac{1}{3}$
b) $\frac{7}{{12}} - \frac{5}{{12}} = \frac{{7 - 5}}{{12}} = \frac{2}{{12}} = \frac{1}{6}$
c) $\frac{{17}}{{21}} - \frac{{10}}{{21}} = \frac{{17 - 10}}{{21}} = \frac{7}{{21}} = \frac{1}{3}$
lấy (1/3 + 1/15 +1/10 + 1/21 ) + (1/36 + 1/28 + 1/6) + (1/45 + 1/55)
= (4/50 + 3/70) + 2/100
= 7/120 + 2/100
= 9/220
Cách giải
X = 8/9 / ( 1+1/3+1/6+1/10+1/15+1/21)
1+1/3=4/3
4/3+1/6=3/2
3/2+1/10=8/5
8/5+1/15=5/3
5/3+1/21=12/7
4/3+3/2=17/6
8/5+5/3=49/15
(17/6+49/15)+17/6=183/30+17/6=286/30
8/9:286/30=2288/270=1140/135=228/27=76/9
X = 76/9
a) $\frac{1}{6} + \frac{3}{2} + \frac{1}{2} = \frac{1}{6} + \left( {\frac{3}{2} + \frac{1}{2}} \right) = \frac{1}{6} + \frac{4}{2} = \frac{1}{6} + \frac{{12}}{6} = \frac{{13}}{6}$
b) $\frac{3}{8} + \frac{1}{2} + \frac{1}{8} = \left( {\frac{3}{8} + \frac{1}{8}} \right) + \frac{1}{2} = \frac{1}{2} + \frac{1}{2} = \frac{2}{2} = 1$
c) $\frac{2}{5} + \frac{6}{{10}} + \frac{3}{5} = \frac{2}{5} + \frac{3}{5} + \frac{3}{5} = \frac{{2 + 3 + 3}}{5} = \frac{8}{5}$
9/1-1/90-1/72-1/56-1/42-1/30-1/20-1/12-1/6-1/2=0/4
Giải :
ta có
\(\frac{9}{10}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
=\(\frac{9}{10}-\left(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{9\times10}\right)\)
=\(\frac{9}{10}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
=\(\frac{9}{10}-\left[1+\left(\frac{-1}{2}+\frac{1}{2}\right)+\left(\frac{-1}{3}+\frac{1}{3}\right)+...+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{10}\right]\)
=\(\frac{9}{10}-\left(1-\frac{1}{10}\right)\)
=\(\frac{9}{10}-1+\frac{1}{10}=0\) (Mong online math ks cho mình nhé)
a) $\frac{1}{{10}} + \frac{3}{{10}} = \frac{{1 + 3}}{{10}} = \frac{4}{{10}} = \frac{2}{5}$
b) $\frac{5}{{12}} + \frac{1}{{12}} = \frac{{5 + 1}}{{12}} = \frac{6}{{12}} = \frac{1}{2}$
c) $\frac{3}{2} + \frac{1}{2} = \frac{{3 + 1}}{2} = \frac{4}{2} = 2$
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) \(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{41\cdot43}=\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+\frac{9-7}{7\cdot9}+...+\frac{43-41}{41\cdot43}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{41}-\frac{1}{43}=\frac{1}{3}-\frac{1}{43}=\frac{43-3}{3\cdot43}=\frac{40}{129}.\)
Ta có:
\(\left(\frac{1}{21}+\frac{1}{210}+\frac{1}{2010}\right)\)\(\times\)\(\left(\frac{1}{3}-\frac{1}{30}-\frac{1}{5}-\frac{1}{10}\right)\)
= \(\left(\frac{1}{21}+\frac{1}{210}+\frac{1}{2010}\right)\)\(\times\)\(\left(\frac{10}{30}-\frac{1}{30}-\frac{6}{30}-\frac{3}{30}\right)\)
= \(\left(\frac{1}{21}+\frac{1}{210}+\frac{1}{2010}\right)\)\(\times\)\(\left(\frac{10-1-6-3}{30}\right)\)
= \(\left(\frac{1}{21}+\frac{1}{210}+\frac{1}{2010}\right)\)\(\times\)\(0\)
= \(0\)
ta có \(\frac{1}{3}-\frac{1}{30}-\frac{1}{5}-\frac{1}{10}=\frac{10-1-6-3}{30}=\frac{0}{30}=0\)
=>\(\left(\frac{1}{21}+\frac{1}{210}+\frac{1}{2010}\right)x\left(\frac{1}{3}-\frac{1}{30}-\frac{1}{5}-\frac{1}{10}\right)=0\)