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chia hết vì những số có chữ số tận cùng là số chẵn thì chia hết cho 2 _ quy tắc chia hết cho 2
a)1/3+1/4+2/3+3/4
=(1/3+2/3)+(1/4+3/4)
=1+1
=2.
b)1/2+1/3-1/5+1/6
=(1/2+1/3+1/6)-1/5
=1-1/5
=4/5
c)2/3x4/5+1/3x4/5
=4/5x(2/3+1/3)
=4/5x1
=4/5
d)2/3x4/5-1/3x4/5
=4/5x(2/3-1/3)
=4/5x1/3
=4/15
a, \(2\dfrac{1}{3}+4\dfrac{1}{5}+4\dfrac{1}{3}\)
\(=\dfrac{7}{3}+\dfrac{21}{5}+\dfrac{13}{3}\)
\(=\dfrac{1}{3}\left(7+13\right)+\dfrac{21}{5}\)
\(=\dfrac{20}{3}+\dfrac{21}{5}=\dfrac{100+63}{15}=\dfrac{163}{15}\)
b, \(5\dfrac{3}{4}-4\dfrac{1}{2}.3\dfrac{7}{8}\)
\(=\dfrac{23}{4}-\dfrac{9}{2}.\dfrac{31}{8}\)
\(=\dfrac{23}{4}-\dfrac{279}{16}=\dfrac{92-279}{16}=-\dfrac{187}{16}\)
c, \(1\dfrac{1}{2}.3\dfrac{2}{3}.4\dfrac{3}{4}\)
\(=\dfrac{3}{2}.\dfrac{11}{3}.\dfrac{19}{4}=\dfrac{209}{8}\)
d, \(6\dfrac{4}{5}:2\dfrac{3}{4}:1\dfrac{1}{2}\)
\(=\dfrac{34}{5}:\dfrac{11}{4}:\dfrac{3}{2}\)
\(=\left(\dfrac{34}{5}.\dfrac{4}{11}\right).\dfrac{2}{3}=\dfrac{136}{55}.\dfrac{2}{3}=\dfrac{272}{165}\)
[4\(\dfrac{1}{5}\) - 2\(\dfrac{2}{5}\)] x 8\(\dfrac{5}{6}\)
= [\(\dfrac{21}{5}\) - \(\dfrac{12}{5}\)] x \(\dfrac{53}{6}\)
= \(\dfrac{9}{5}\) x \(\dfrac{53}{6}\)
= \(\dfrac{159}{10}\)
[5\(\dfrac{1}{3}\) - 2\(\dfrac{2}{3}\)]: \(\dfrac{3}{7}\)
= [\(\dfrac{16}{3}\) - \(\dfrac{8}{3}\)]: \(\dfrac{3}{7}\)
= \(\dfrac{8}{3}\) : \(\dfrac{3}{7}\)
= \(\dfrac{56}{9}\)
\(=\frac{9}{7}+\frac{23}{6}+\frac{23}{4}-\frac{63}{562}\)x \(\frac{205}{532}:\frac{41}{3}\)
\(=\frac{215}{42}+\frac{23}{4}-\frac{1845}{42712}:\frac{41}{3}\)
\(=\frac{913}{84}-\frac{135}{42712}\)
\(=10,86588691...\)
(1+2):3=1
1x2+3-4=1
1-2+3+4-5=1
(1x2+3-4+5):6=1
(1-2+3+4-5+6):7=1
[(1x2+3-4+5):6+7]:8=1
a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)
b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)
`1 + 2 + 3 + 4 + 5 +...+ 1000000000000`
khoảng cách : `1`
Số số hạng là : `(1000000000000-1) :1 +1=1000000000000`
Tổng là : `(1000000000000+1 ) xx 1000000000000 : 2=500000 000000 500000 000000 `
1 + 2 + 3 + 4 + 5 +...+ 1000000000000
=1000000000000*(1000000000000-1)/2