1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1-18 = ?
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\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times...\times\left(1-\frac{1}{18}\right)\times\left(1-\frac{1}{19}\right)\times\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{17}{18}\times\frac{18}{19}\times\frac{19}{20}\)
\(=\frac{1\times2\times3\times...\times17\times18\times19}{2\times3\times4\times...\times18\times19\times20}\)
\(=\frac{1}{20}\)
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{19}\right).\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.....\frac{19}{20}\)
\(=\frac{1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.17.18.19}{2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.17.18.19.20}\)(RÚT GỌN CÁC SỐ GIỐNG NHAU)
\(=\frac{1}{20}\)
\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}{\dfrac{19}{1}+\dfrac{18}{2}+\dfrac{17}{3}+....+\dfrac{1}{19}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}{1+\left(\dfrac{18}{2}+1\right)+\left(\dfrac{17}{3}+1\right)+\left(\dfrac{1}{19}+1\right)}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}{1+\dfrac{20}{2}+\dfrac{20}{3}+...+\dfrac{20}{19}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}}{20.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}\right)}\)
\(=\dfrac{1}{20}\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{20}\right)=\frac{1}{2}.\frac{2}{3}...\frac{19}{20}=\frac{1.2...19}{2.3...20}=\frac{1}{20}\)
\(\left(1+\frac{1}{15}\right).\left(1+\frac{1}{16}\right).\left(1+\frac{1}{17}\right).\left(1+\frac{1}{18}\right)\)
=\(\frac{16}{15}.\frac{17}{16}.\frac{18}{17}.\frac{19}{18}\)
=\(\frac{19}{15}\)
Xét tử số
1/18+2/17+3/16+...+18/1+18
=[(1/18)+1]+[(2/17)+1]+[(3/16)+1]+...+[(18/1)+1]
=19/18+19/17+19/16+...+19/1
=19.[(1/18)+(1/17)+(1/16)+...+1/1]
=>phân số trên bằng 19
1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1-18 = 0
TL
theo như mình tính được là 0
nha
_HT_