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\(C=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)......\left(1-\frac{1}{100}\right).\)
<=> \(C=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{99}{100}\)
<=> \(C=\frac{1.2.3....99}{2.3.4....100}\)
<=> \(C=\frac{1}{100}\)
\(A=4+\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{19}\right)\cdot\left(1-\frac{1}{20}\right)\)
\(A=4+\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{18}{19}\cdot\frac{19}{20}\right)\)
\(A=4+\frac{1\cdot2\cdot3\cdot...\cdot18\cdot19}{2\cdot3\cdot4\cdot...\cdot19\cdot20}\)
\(A=4+\frac{1}{20}\)
\(A=\frac{81}{20}\)
\(a)\) \(427-98=329\)
\(b)\) \(2\cdot19\cdot15+3\cdot43\cdot10+62\cdot80\)
\(=\left(2\cdot15\right)\cdot19+\left(3\cdot10\right)\cdot43+62\cdot80\)
\(=30\cdot19+30\cdot43+62\cdot80\)
\(=30\cdot\left(19+43\right)+62\cdot80\)
\(=30\cdot62+62\cdot80\)
\(=62\cdot\left(30+80\right)\)
\(=62\cdot110=6820\)
\(c)\) Đặt \(M=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\)
\(\Rightarrow3M=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
\(\Rightarrow3M-M=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\right)\)
\(\Rightarrow2M=1-\frac{1}{3^6}\)
\(\Rightarrow M=\frac{728}{2\cdot729}=\frac{364}{729}\)
Vậy \(M=\frac{364}{729}\)
B=(1-\(\frac{1}{2}\)).(1-\(\frac{1}{3}\)).(1-\(\frac{1}{4}\)).....(1-\(\frac{1}{20}\))
B=\(\frac{1}{2}\) . \(\frac{2}{3}\) . \(\frac{3}{4}\) .... \(\frac{19}{20}\)
B=\(\frac{1.2.3....19}{2.3.4....20}\)
B=\(\frac{1}{20}\)(mk rút gọn nha:2 với 2;3 với 3;4 với 4;....;19 với 19;còn lại là\(\frac{1}{20}\))
So sánh:\(\frac{1}{21}\) và \(\frac{1}{20}\)
Quy đồng:\(\frac{20}{420}\) và \(\frac{21}{420}\)
Vì 20<21 =>\(\frac{1}{21}\) <\(\frac{1}{20}\)
\(\left(1-\frac{1}{3}\right).\left(1-\frac{1}{6}\right).\left(1-\frac{1}{10}\right).\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{780}\right).a=1\)
\(\left(\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}...\frac{779}{780}\right).a=1\)
\(\left(\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.\frac{28}{30}...\frac{1558}{1560}\right).a=1\)
\(\left(\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.\frac{4.7}{5.6}...\frac{38.41}{39.40}\right).a=1\)
\(\left(\frac{1.2.3.4...38}{3.4.5.6..40}.\frac{4.5.6.7...41}{2.3.4.5..39}\right).a=1\)
\(\left(\frac{2}{39.40}.\frac{40.41}{2.3}\right).a=1\)
\(\frac{41}{39.3}.a=1\)
\(\frac{41}{117}.a=1\)
\(a=1:\frac{41}{117}\)
\(a=1.\frac{117}{41}=\frac{117}{41}\)
Vậy a = 117/41
Ủng hộ mk nha ^_-
\(=\left[\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{9998}{9999}\right]\cdot\frac{1999}{2000}=\frac{1\cdot2\cdot3\cdot4\cdot...\cdot9998}{2\cdot3\cdot4\cdot5\cdot...\cdot9999}\cdot\frac{1999}{2000}=\frac{1}{9999}\cdot\frac{1999}{2000}=\frac{1}{2000}\)
=\(\frac{1}{2}\). \(\frac{2}{3}\).\(\frac{3}{4}\)... \(\frac{1999}{2000}\)
=\(\frac{1}{2}\)- \(\frac{1999}{2000}\)
= \(\frac{-999}{2000}\)
