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\(\dfrac{1}{20}=\dfrac{1}{4x5}=\dfrac{1}{4}-\dfrac{1}{5}\)
Tương tự các phân số khác
S= \(\dfrac{1}{4}-\dfrac{1}{12}=\dfrac{1}{6}\)
\(\dfrac{1}{20}+\dfrac{1}{30}\)+ \(\dfrac{1}{42}\)+\(\dfrac{1}{56}\)+\(\dfrac{1}{72}\)+\(\dfrac{1}{90}\)+\(\dfrac{1}{110}\)+\(\dfrac{1}{132}\)
= \(\dfrac{1}{4\times5}\)+\(\dfrac{1}{5\times6}\)+\(\dfrac{1}{6\times7}\)+\(\dfrac{1}{7\times8}\)+\(\dfrac{1}{8\times9}\)+\(\dfrac{1}{9\times10}\)+\(\dfrac{1}{10\times11}\)+\(\dfrac{1}{11\times12}\)
= \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)+\(\dfrac{1}{9}\)-\(\dfrac{1}{10}\)+\(\dfrac{1}{10}\)-\(\dfrac{1}{11}\)+\(\dfrac{1}{11}\)-\(\dfrac{1}{12}\)
= \(\dfrac{1}{4}\) - \(\dfrac{1}{12}\)
= \(\dfrac{3}{12}\) - \(\dfrac{1}{12}\)
= \(\dfrac{2}{12}\)
=\(\dfrac{1}{6}\)
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`=1/[4xx5]+1/[5xx6]+1/[6xx7]+...+1/[11xx12]`
`=1/4-1/5+1/5-1/6+1/6-1/7+...+1/11-1/12`
`=1/4-1/12=3/12-1/12=2/12=1/6`
\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\\ =\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}+\dfrac{1}{8\times9}+\dfrac{1}{9\times10}+\dfrac{1}{10\times11}+\dfrac{1}{11\times12}\\ =\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\\ =\dfrac{1}{4}-\dfrac{1}{12}\\ =\dfrac{3}{12}-\dfrac{1}{12}=\dfrac{2}{12}=\dfrac{1}{6}\)
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a) 5/30+15/90+25/150+35/210+45/270
=1/6+1/6+1/6+1/6+1/6
=1/6 x 5
=5/6
b) 1/2+1/6+1/12+1/20+....+1/56
=1/1x2+1/2x3+1/3x4+1/4x5+.....1/7x8
=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+.......-1/7+1/7-1/8
=1/1-1/8
=7/8
c) mình chịu
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\(\frac{3}{12}+\frac{3}{20}+\frac{3}{30}+...+\frac{3}{110}\)
\(=3\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{110}\right)\)
\(=3\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{10\cdot11}\right)\)
\(=3\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=3\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=3\cdot\frac{8}{33}=\frac{8}{11}\)
https://olm.vn/hoi-dap/detail/63463423750.html
vào tham khảo đi mk nhác trình bày
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\(=\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{1}{12}\right)+...+\left(1-\dfrac{1}{90}\right)\\ =\left(1+1+1+1+1+1+1+1+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)\\ =9-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\right)\\ =9-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ =9-\left(1-\dfrac{1}{10}\right)=9-\dfrac{9}{10}=\dfrac{81}{10}\)
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a) \(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{6}+1-\frac{1}{12}+...+1-\frac{1}{90}\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\right)\)
Từ 2 đến 9 có : ( 9 - 2 ) / 1 + 1 = 8 ( số hạng ) => có 8 số 1
\(\Rightarrow8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=8-\frac{2}{5}=\frac{38}{5}\)
b) \(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{109}{110}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+...+1-\frac{1}{110}\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{10\cdot11}\right)\)
Từ 1 đến 10 có : ( 10 - 1 ) / 1 + 1 = 10 ( số hạng ) => có 10 số 1
\(\Rightarrow10-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=10-\left(1-\frac{1}{11}\right)\)
\(=10-\frac{10}{11}=\frac{100}{11}\)
ko tính được 90 :56 làm sao
90:45+90:56+90=90:(45+56+1)=90:102=15/17