D=(\(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{780}\right)\)
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\(B=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{780}\right)\)
\(\Rightarrow B=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}...\frac{779}{780}\)
\(\Rightarrow B=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.\frac{28}{30}...\frac{1558}{1560}\)
\(\Rightarrow B=\frac{1.4}{2.3}.\frac{2.5}{3.4}\frac{3.6}{4.5}...\frac{38.41}{39.40}\)
\(\Rightarrow B=\frac{\left(1.2.3...38\right)\left(4.5.6...41\right)}{\left(2.3.4...39\right)\left(3.4.5...40\right)}\)
\(\Rightarrow B=\frac{1.41}{39.3}=\frac{41}{117}\)
Vậy B=\(\frac{41}{117}\)
Ai thấy đúng thì k nha
\(\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{6}\right)\cdot\cdot\cdot\left(1-\frac{1}{780}\right)\)
\(=\frac{2}{3}\cdot\frac{5}{6}\cdot\cdot\cdot\frac{779}{780}\)
\(=\frac{4}{6}\cdot\frac{10}{12}\cdot\cdot\cdot\frac{1578}{1560}\)
\(=\frac{1\cdot4}{2\cdot3}\cdot\frac{2\cdot5}{3\cdot4}\cdot\cdot\cdot\frac{38\cdot41}{39\cdot40}\)
\(=\frac{\left(1\cdot4\right)\cdot\left(2\cdot5\right)\cdot\cdot\cdot\left(38\cdot41\right)}{\left(2\cdot3\right)\cdot\left(3\cdot4\right)\cdot\cdot\cdot\left(39\cdot40\right)}\)
\(=\frac{\left(1\cdot2\cdot\cdot\cdot38\right)\cdot\left(4\cdot5\cdot\cdot\cdot41\right)}{\left(2\cdot3\cdot\cdot\cdot39\right)\cdot\left(3\cdot4\cdot\cdot\cdot40\right)}\)
\(=\frac{1\cdot41}{39\cdot3}\)
\(=\frac{41}{117}\)
\(\left(1-\frac{1}{3}\right).\left(1-\frac{1}{6}\right).\left(1-\frac{1}{10}\right).\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{780}\right).a=1\)
\(\left(\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}...\frac{779}{780}\right).a=1\)
\(\left(\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.\frac{28}{30}...\frac{1558}{1560}\right).a=1\)
\(\left(\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}...\frac{38.41}{39.40}\right).a=1\)
\(\left(\frac{1.2.3.4...38}{3.4.5.6..40}.\frac{4.5.6.7...41}{2.3.4.5..39}\right).a=1\)
\(\left(\frac{2}{39.40}.\frac{40.41}{2.3}\right).a=1\)
\(\frac{41}{39.3}.a=1\)
\(\frac{41}{117}.a=1\)
\(a=1:\frac{41}{117}\)
\(a=1.\frac{117}{41}=\frac{117}{41}\)
Vậy a = 117/41
Ủng hộ mk nha ^_-
D = $\frac{2}{3}.\frac{5}{6}.\frac{9}{10}. ... .\frac{799}{780}$
= $\frac{2.2}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}. ... .\frac{38.41}{39.40}$
= $\frac{2.2}{2.3}.\frac{2.3. ... .38}{3.4. ... 39}.\frac{5.6. ... .41}{4.5. ... .40}$
= $\frac{2}{3}.\frac{2}{39}.\frac{41}{4}$
= $\frac{41}{3.39}$
D = \(\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}.....\frac{779}{780}\)
= \(\frac{2.2}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}.....\frac{38.41}{39.40}\)
= \(\frac{2}{3}.\frac{2.3.4....38}{3.4.5....39}.\frac{5.6.7.....41}{4.5.6.....40}\)
= \(\frac{2}{3}.\frac{2}{39}.\frac{41}{4}\)
= \(\frac{41}{117}\)
\(D=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{780}\right)\)
\(D=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}...\frac{779}{780}\)
\(D=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.\frac{28}{30}...\frac{1558}{1560}\)
\(D=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}...\frac{38.41}{39.40}\)
\(D=\frac{1.2.3.4...38}{3.4.5.6...40}.\frac{4.5.6.7...41}{2.3.4.5...39}\)
\(D=\frac{2}{39.40}.\frac{40.41}{2.3}\)
\(D=\frac{41}{39.3}=\frac{41}{117}\)
A=(−2/3 ).(−5/6 ).(−9/10 )...(−779/780 )=(−4/6 ).(−10/12 ).(−18/20 )....(−1558/1560 )
A=(−1.4).(−2.5).(−3.6)....(−38.41) / (2.3).(3.4).(4.5)....(39.40) =(1.2.3....38).(4.5.6...41) / (2.3.4...39).(3.4.5...40) ( Vì từ -1 đến -38 có 38 số =>tích của 38 số âm = tích của 38 số dương)
A=(1.2.3....38).(4.5.6...41) / (2.3.4...39).(3.4.5...40) =1.41/39.3 =41/ 117