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\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}=\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\\ =\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\\ =\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+\left(-\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(-\dfrac{1}{5}+\dfrac{1}{5}\right)+\left(-\dfrac{1}{6}+\dfrac{1}{6}\right)+\left(-\dfrac{1}{7}+\dfrac{1}{7}\right)+\left(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{8}\right)\\ =\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{8}=1-\dfrac{1}{8}=\dfrac{8-1}{8}=\dfrac{7}{8}\)
`@` `\text {Ans}`
`\downarrow`
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
`=`\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}\)
`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
`=`\(1-\dfrac{1}{8}\)
`=`\(\dfrac{7}{8}\)
A = \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\)+...+ \(\dfrac{1}{812}\) + \(\dfrac{1}{870}\)
A = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\)+...+ \(\dfrac{1}{28\times29}\)+ \(\dfrac{1}{29\times30}\)
A = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +...+\(\dfrac{1}{28}\)-\(\dfrac{1}{29}\)+ \(\dfrac{1}{29}\) - \(\dfrac{1}{30}\)
A = 1 - \(\dfrac{1}{30}\)
A = \(\dfrac{29}{30}\)
1/2+1/6+1/12+1/20+...+1/380
=1/1.2+ 1/2.3+ 1/3.4+ 1/4.5+...+ 1/19.20
= 1/1-1/2+ 1/2-1/3+ 1/3-1/4+ 1/4-1/5+...+1/19-1/20
=1-1/29
=29/29-1/29
= 28/29
\(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{110}\)
= \(\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{10\times11}\)
= \(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{10}-\dfrac{1}{11}\)
= \(\dfrac{1}{3}-\dfrac{1}{11}=\dfrac{8}{33}\)
\(a.32,5-3\cdot0,87=32,5-2,61=29,89\)
\(8,5\cdot\left(1\dfrac{1}{2}+\dfrac{4}{4}\right):5=8,5\cdot\left(\dfrac{3}{2}+\dfrac{4}{4}\right):5\\ =8,5\cdot\left(\dfrac{6}{4}+\dfrac{4}{4}\right):5\\ =8,5\cdot\dfrac{10}{4}:5\\ =\dfrac{85}{4}:5\\ =\dfrac{17}{4}\)
\(b.30,96-6,45+14,4:3=30,96-6,45+4,8\\ =29,31\)
\(\dfrac{2}{5}\cdot\left(\dfrac{4}{5}-\dfrac{1}{2}\right)=\dfrac{2}{5}\cdot\left(\dfrac{8}{10}-\dfrac{5}{10}\right)\\ =\dfrac{2}{5}\cdot\dfrac{3}{10}=\dfrac{3}{25}\)
bài 2
\(a.2,5\cdot12,5\cdot8\cdot0,4=\left(2,5\cdot0,4\right)\left(12,5\cdot8\right)\\ =1\cdot100=100\)
b,\(\dfrac{12}{15}\cdot\dfrac{5}{6}\cdot\dfrac{3}{20}\cdot\dfrac{32}{5}=\dfrac{12\cdot5\cdot3\cdot32}{15\cdot6\cdot20\cdot5}\\ =\dfrac{3\cdot4\cdot5\cdot3\cdot4\cdot8}{3\cdot5\cdot2\cdot3\cdot5\cdot4\cdot5}=\dfrac{16}{25}\)
Bài 1:
a) \(32.5-3\cdot0.87=32.5-2.61=29.89\)
\(8.5\cdot\left(1\dfrac{1}{2}+\dfrac{4}{4}\right):5=8.5\cdot\dfrac{5}{2}:5=\dfrac{17}{2}\cdot\dfrac{5}{2}:5=\dfrac{85}{4}\cdot\dfrac{1}{5}=\dfrac{17}{4}\)
a) =8 phần 27 ×( 7 phần 4 - 3 phần 8 + 1 phần 2)
= 8 phần 27 × ( 14 phần 8 - 3 phần 8 + 4 phần 8)
= 8 phần 27 × 15 phần 8
= 5 phần 9
b) = 5 phần 18 × ( 2 phần 5 - 1 phần 3 + 5 phần 6)
= 5 phần 18 × ( 12 phần 30 - 10 phần 30 + 25 phần 30)
= 5 phần 18 × 27 phần 30
= 5 phần 18 × 9 phần 10
= 1 phần 4
c) =( 29 phần 10 - 3 phần 10 + 1 phần 3) ÷ 8 phần 15
=( 87 phần 30 - 9 phần 10 + 10 phần 30) ÷ 8 phần 15
= 88 phần 30 × 15 phần 8
= 44 phần 15 × 15 phần 8
= 44 phần 8
= 11 phần 2
d) = 238 ÷ ( 1 phần 2 + 7 phần 4 - 1 phần 4)
= 238 ÷ ( 2 phần 4 + 7 phần 4 - 1 phần 4)
= 238 ÷ 8 phần 4
= 238 × 4 phần 8
= 238 × 1 phần 2
= 119
=1/1x2+1/2x3+1/3x4+1/4x5+1/5x6+1/6x7
=1-1/7
=6/7
\(\frac{113}{140}\frac{ }{ }\)