Tính :
D=\(\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right)..........\left(1-\frac{1}{n^2}\right)\)
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tính các tích sau với nEN, n lớn hơn bằng 2a)left(1-frac{1}{2}right)left(1-frac{1}{3}right)left(1-frac{1}{4}right)...left(1-frac{1}{n}right)b)left(1+frac{1}{2}right)left(1+frac{1}{3}right)left(1+frac{1}{4}right)...left(1-frac{1}{n}right)c)left(1-frac{1}{2^2}right)left(1-frac{1}{3^2}right)left(1-frac{1}{4^2}right)...left(1-frac{1}{n^2}right)
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tính các tích sau với nEN, n lớn hơn bằng 2
a)\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{n}\right)\)
b)\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1-\frac{1}{n}\right)\)
c)\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\)
1, Tính \(\frac{1}{2}-\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)-\left(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right)+...+\left(\frac{1}{100}+\frac{2}{100}+\frac{3}{100}+...+\frac{99}{100}\right)\)2,Tính \(\left(1-\frac{1}{2^2}\right)x\left(1-\frac{1}{3^2}\right)x\left(1-\frac{1}{4^2}\right)x...x\left(1-\frac{1}{n^2}\right)\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{n+1}\right)\left(n\in N\right)\)
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+.......+\frac{1}{20}\left(1+2+3+4....+20\right)\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{n+1}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{n}{n+1}\)
\(=\frac{1}{n+1}\)
\(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)...+\frac{1}{20}.\left(1+2+3+...+20\right)\)
\(=1+\frac{1}{2}.2.3:2+\frac{1}{3}.3.4:2+\frac{1}{4}.4.5:2+...+\frac{1}{20}.20.21:2\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{21}{2}\)
\(=\frac{2+3+4+5+...+21}{2}=115\)
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Bài 2a) Aleft(frac{1}{2}-1right)left(frac{1}{3}-1right)...left(frac{1}{2002}-1right)left(frac{1}{2003}-1right)b) Bleft(-1frac{1}{2^2}right)left(-1frac{1}{3^2}right)left(-1frac{1}{4^2}right)...left(-1frac{1}{2003^2}right)left(-1frac{1}{2004^2}right)c) Cleft(1-frac{1}{2^2}right)left(1-frac{1}{3^2}right)left(1-frac{1}{4^2}right)...left(1-frac{1}{n^2}right)left(nin N,nge2right)
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Bài 2
a) \(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{2002}-1\right)\left(\frac{1}{2003}-1\right)\)
b) \(B=\left(-1\frac{1}{2^2}\right)\left(-1\frac{1}{3^2}\right)\left(-1\frac{1}{4^2}\right)...\left(-1\frac{1}{2003^2}\right)\left(-1\frac{1}{2004^2}\right)\)
c) \(C=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\left(n\in N,n\ge2\right)\)
\(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{2002}-1\right)\left(\frac{1}{2003}-1\right)\)
\(=\left(-\frac{1}{2}\right)\left(-\frac{2}{3}\right)...\left(-\frac{2001}{2002}\right)\left(-\frac{2002}{2003}\right)\)
\(=\frac{-1.\left(-2\right).....\left(-2001\right)\left(-2002\right)}{2.3....2002.2003}\)
\(=\frac{1}{2003}\)
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Tính các tích sau: với n là số tự nhiên, n<3
a) \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{n}\right)\)
b) \(\left(1-\frac{1}{2^2}\right)\cdot\left(1-\frac{1}{3^2}\right)\cdot\left(1-\frac{1}{4^2}\right)\cdot...\cdot\left(1-\frac{1}{n^2}\right)\)
help me! (ngu toàn tập)a)frac{3}{left(1.2right)^2}+frac{5}{left(2.3right)^2}+...+frac{2n+1}{left[nleft(n+1right)right]^2}b)left(1-frac{1}{2^2}right)left(1-frac{1}{3^2}right)left(1-frac{1}{4^2}right)....left(1-frac{1}{n^2}right)c)frac{150}{5.8}+frac{150}{8.11}+frac{150}{11.14}+...+frac{150}{47.50}d)frac{1}{1.2.3}+frac{1}{2.3.4}+frac{1}{3.4.5}+...+frac{1}{left(n-1right)nleft(n+1right)}
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help me! (ngu toàn tập)
a)\(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
b)\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)....\left(1-\frac{1}{n^2}\right)\)
c)\(\frac{150}{5.8}+\frac{150}{8.11}+\frac{150}{11.14}+...+\frac{150}{47.50}\)
d)\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\left(n-1\right)n\left(n+1\right)}\)
\(\frac{150}{5.8}+\frac{150}{8.11}+\frac{150}{11.14}+.....+\frac{150}{47.50}\)
\(=50.\left(\frac{3}{5.8}+\frac{5}{8.11}+.....+\frac{3}{47.50}\right)\)
\(=50.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{47}-\frac{1}{50}\right)\)
\(=50.\left(\frac{1}{5}-\frac{1}{50}\right)\)
\(=50.\frac{9}{50}=9\)
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Bài 1 Thưc hiện phép tính ( tính nhanh nếu có thể)a)frac{-1}{24}-left[frac{1}{4}-left(frac{1}{2}-frac{7}{8}right)right]b)left(frac{5}{7}-frac{7}{5}right)-left[frac{1}{2}-left(frac{-2}{7}-frac{1}{10}right)right]C)left(frac{-1}{2}right)-left(frac{-3}{5}right)+left(frac{-1}{9}right)+frac{1}{17}-left(frac{-2}{7}right)+frac{4}{35}-frac{7}{18}d)left(3-frac{1}{4}+frac{2}{3}right)-left(5-frac{1}{3}-frac{6}{5}right)-left(6-frac{7}{4}+frac{3}{2}right)
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Bài 1 Thưc hiện phép tính ( tính nhanh nếu có thể)
a)\(\frac{-1}{24}-\left[\frac{1}{4}-\left(\frac{1}{2}-\frac{7}{8}\right)\right]\)
b)\(\left(\frac{5}{7}-\frac{7}{5}\right)-\left[\frac{1}{2}-\left(\frac{-2}{7}-\frac{1}{10}\right)\right]\)
C)\(\left(\frac{-1}{2}\right)-\left(\frac{-3}{5}\right)+\left(\frac{-1}{9}\right)+\frac{1}{17}-\left(\frac{-2}{7}\right)+\frac{4}{35}-\frac{7}{18}\)
d)\(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
Giúp mik vớiTính nhanh:a. Aleft(-1right)^{2n}.left(-1right)^n.left(-1right)^{n+1}left(nin Nright)b. Bleft(10000-1^2right)left(10000-2^2right)left(10000-3^2right)..left(10000-1000^2right)c. Cleft(frac{1}{125}-frac{1}{1^3}right)left(frac{1}{125}-frac{1}{2^3}right)left(frac{1}{125}-frac{1}{3^3}right)...left(frac{1}{125}-frac{1}{25^3}right)d. D1999^{left(1000-1^3right)left(1000-2^3right)left(1000-3^3right)...left(1000-10^3right)}
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Giúp mik với
Tính nhanh:
a. A=\(\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}\left(n\in N\right)\)
b. B=\(\left(10000-1^2\right)\left(10000-2^2\right)\left(10000-3^2\right)..\left(10000-1000^2\right)\)
c. C=\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)\left(\frac{1}{125}-\frac{1}{3^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
d. D=\(1999^{\left(1000-1^3\right)\left(1000-2^3\right)\left(1000-3^3\right)...\left(1000-10^3\right)}\)
a) \(A=\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}=\left(-1\right)^{3n+1}\)
b) \(B=\left(10000-1^2\right)\left(10000-2^2\right).........\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right)......\left(10000-100^2\right)....\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right).....\left(10000-10000\right).....\left(10000-1000^2\right)=0\)
c) \(C=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)..........\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right).....\left(\frac{1}{125}-\frac{1}{5^3}\right)......\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)........\left(\frac{1}{125}-\frac{1}{125}\right).....\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
d) \(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-10^3\right)}\)
\(=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-1000\right)}=1999^0=1\)
a)Tính: {left( {frac{{ - 1}}{2}} right)^5};{left( {frac{{ - 2}}{3}} right)^4};{left( { - 2frac{1}{4}} right)^3};{left( { - 0,3} right)^5};{left( { - 25,7} right)^0}.b)Tính: {left( { - frac{1}{3}} right)^2};{left( { - frac{1}{3}} right)^3};{left( { - frac{1}{3}} right)^4};{left( { - frac{1}{3}} right)^5}.Hãy rút ra nhận xét về dấu của luỹ thừa với số mũ chẵn và luỹ thừa với số mũ lẻ của một số hữu tỉ âm.
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a)Tính: \({\left( {\frac{{ - 1}}{2}} \right)^5};{\left( {\frac{{ - 2}}{3}} \right)^4};{\left( { - 2\frac{1}{4}} \right)^3};{\left( { - 0,3} \right)^5};{\left( { - 25,7} \right)^0}\).
b)Tính: \({\left( { - \frac{1}{3}} \right)^2};{\left( { - \frac{1}{3}} \right)^3};{\left( { - \frac{1}{3}} \right)^4};{\left( { - \frac{1}{3}} \right)^5}\).
Hãy rút ra nhận xét về dấu của luỹ thừa với số mũ chẵn và luỹ thừa với số mũ lẻ của một số hữu tỉ âm.
a)
\(\begin{array}{l}{\left( {\frac{{ - 1}}{2}} \right)^5} = \frac{{{{\left( { - 1} \right)}^5}}}{{{2^5}}} = \frac{{ - 1}}{{32}};\\{\left( {\frac{{ - 2}}{3}} \right)^4} = \frac{{{{\left( { - 2} \right)}^4}}}{{{3^4}}} = \frac{{16}}{{81}};\\{\left( { - 2\frac{1}{4}} \right)^3} = {\left( {\frac{{ - 9}}{4}} \right)^3} = \frac{{{{\left( { - 9} \right)}^3}}}{{{4^3}}} = \frac{{-729}}{{64}};\\{\left( { - 0,3} \right)^5} = {\left( {\frac{{ - 3}}{{10}}} \right)^5} = \frac{{ - 243}}{{100000}};\\{\left( { - 25,7} \right)^0} = 1\end{array}\)
b)
\(\begin{array}{l}{\left( { - \frac{1}{3}} \right)^2} = \frac{1}{9};\\{\left( { - \frac{1}{3}} \right)^3} = \frac{{ - 1}}{{27}};\\{\left( { - \frac{1}{3}} \right)^4} = \frac{1}{{81}};\\{\left( { - \frac{1}{3}} \right)^5} = \frac{{ - 1}}{{243}}.\end{array}\)
Nhận xét:
+ Luỹ thừa của một số hữu tỉ âm với số mũ chẵn là một số hữu tỉ dương.
+ Luỹ thừa của một số hữu tỉ âm với số mũ lẻ là một số hữu tỉ âm.
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