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\(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).............\left(1-\dfrac{1}{2014}\right)\)
\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right).\left(\dfrac{3}{3}-\dfrac{1}{3}\right)..............\left(\dfrac{2014}{2014}-\dfrac{1}{2014}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}................\dfrac{2013}{2014}\)
\(=\dfrac{1}{2014}\)
\(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right)....\left(1-\dfrac{1}{2014}\right)\)
\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right).\left(\dfrac{3}{3}-\dfrac{1}{3}\right).....\left(\dfrac{2014}{2014}-\dfrac{1}{2014}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.....\dfrac{2013}{2014}\)
\(=\dfrac{1}{2014}\)
Chúc bạn học tốt!
Ta có:
\(\left(1-\dfrac{1}{2^2}\right).\left(1-\dfrac{1}{3^2}\right).\left(1-\dfrac{1}{4^2}\right).....\left(1-\dfrac{1}{n^2}\right)\)
\(=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}.....\dfrac{n^2-1}{n^2}\)
\(=\dfrac{3.8.15....\left(n^2-1\right)}{4.9.16.....n^2}\)
\(=\dfrac{1.3.2.4.3.5....\left(n-1\right)\left(n+1\right)}{2.2.3.3.4.4....n.n}\)
\(=\dfrac{\left[1.2.3....\left(n-1\right)\right].\left[3.4.5....\left(n+1\right)\right]}{\left(2.3.4....n\right).\left(2.3.4....n\right)}\)
\(=\dfrac{1.\left(n+1\right)}{n.2}=\dfrac{n+1}{2n}\)
Ta có công thức:
\(1-\dfrac{1}{k^2}=\dfrac{k^2-1^2}{k^2}=\dfrac{\left(k+1\right)\left(k+2\right)}{k^2}\)
Áp dụng công thức trên ta đc:
\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)....\left(1-\dfrac{1}{n^2}\right)\)
\(=\dfrac{2^2-1^2}{2^2}.\dfrac{3^2-1^2}{3^2}.\dfrac{4^2-1^2}{4^2}....\dfrac{n^2-1^2}{n^2}\)
\(=\dfrac{\left(2+1\right)\left(2-1\right)}{2.2}.\dfrac{\left(3+1\right)\left(3-1\right)}{3.3}.\dfrac{\left(4+1\right)\left(4-1\right)}{4.4}....\dfrac{\left(n+1\right)\left(n-1\right)}{n.n}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}....\dfrac{\left(n+1\right)\left(n-1\right)}{n.n}\)
\(=\dfrac{[1.2.3....\left(n+1\right)].[3.4.5....\left(n-1\right)]}{\left(2.3.4....n\right)\left(2.3.4....n\right)}\)
\(=\left(n+1\right).\dfrac{1}{2n}=\dfrac{n+1}{2n}\)
Chúc bạn học tốt!
\(B=\left(\frac{3}{5}\right)^2\cdot5^2-\left(2\frac{1}{4}\right)^3:\left(\frac{3}{4}\right)^3+\frac{1}{2}\)
\(B=\left(\frac{3}{5}\cdot5\right)^2-\left(\frac{9}{4}:\frac{3}{4}\right)^3+\frac{1}{2}\)
\(B=3^2-\left(\frac{9}{4}\cdot\frac{4}{3}\right)^3+\frac{1}{2}\)
\(B=3^2-3^3+\frac{1}{2}=-18+\frac{1}{2}=-\frac{35}{2}\)
Ta có \(-A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{2014^2}\right)\)
\(=\left(\frac{2^2-1}{2^2}\right)\left(\frac{3^2-1}{3^2}\right)...\left(\frac{2014^2-1}{2014^2}\right)\)
\(=\frac{\left(2-1\right)\left(2+1\right)}{2^2}.\frac{\left(3-1\right)\left(3+1\right)}{3^2}...\frac{\left(2014-1\right)\left(2014+1\right)}{2014^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{2013.2015}{2014.2014}\)
\(=\frac{1.2...2013}{2.3...2014}.\frac{3.4...2015}{2.3...2014}\)
\(=\frac{1}{2014}.\frac{2015}{2}\)
\(=\frac{2015}{2014.2}>\frac{1}{2}\)hay -A>1/2
=>\(A< \frac{-1}{2}\)hay A<B
a, \(\dfrac{2\cdot8^4\cdot27^2+4\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)
=\(\dfrac{2\cdot\left(2^3\right)^4\cdot\left(3^3\right)^2+2^2\cdot2^9\cdot3^9}{2^7\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot\left(3^2\right)^4}\)
=\(\dfrac{2\cdot2^{12}\cdot3^6+2^{11}\cdot3^9}{2^{14}\cdot3^7+2^{10}\cdot5\cdot3^8}\)
=\(\dfrac{2^{11}\cdot3^6\cdot\left(2^2+3^3\right)}{2^{10}\cdot3^7\cdot\left(2^4+5\cdot3\right)}\)
=\(\dfrac{2^{11}\cdot3^6\cdot31}{2^{10}\cdot3^7\cdot31}\)
=\(\dfrac{2}{3}\)
b, \(\dfrac{\dfrac{8}{27}\cdot\dfrac{9}{16}\cdot\left(-1\right)}{\dfrac{4}{25}\cdot\dfrac{-125}{1728}}\)
=\(\dfrac{\dfrac{8\cdot9\cdot\left(-1\right)}{27\cdot16}}{\dfrac{4\cdot\left(-125\right)}{25\cdot1728}}\)
=\(\dfrac{\dfrac{-1}{6}}{\dfrac{-5}{432}}\)
=\(\dfrac{-1}{6}\cdot\dfrac{-432}{5}\)
=\(\dfrac{72}{5}\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...........\left(1-\frac{1}{2014}\right)\)
\(=\left(\frac{2}{2}-\frac{1}{2}\right)\left(\frac{3}{3}-\frac{1}{3}\right).........\left(\frac{2014}{2014}-\frac{1}{2014}\right)\)
\(=\frac{1}{2}.\frac{2}{3}............\frac{2013}{2014}\)
\(=\frac{1}{2014}\)
(1-1/2).(1-1/3)......(1-1/2014)
=1/2.2/3.....2013/2014
=1.2....2013/2.3.....2014
=1/2014