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\(C=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{99.100.101}\)
\(C=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+....+\frac{101-99}{99.100.101}\)
\(C=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{99.100}+\frac{2}{100.101}\)
\(C=\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
\(C=\frac{1}{2}\cdot\left(\frac{1}{1.2}-\frac{1}{100.101}\right)\)
\(C=\frac{1}{2}\cdot\frac{5049}{10100}=\frac{5049}{20200}\)
Bài này hơi dài nên bạn tham khảo tại đây nha :
Câu hỏi của Kim Sura xXx pÉ heO - Toán lớp 6 - Học toán với OnlineMath
A = \(\frac{1}{1.2.3}\) + \(\frac{1}{2.3.4}\) + ...+ \(\frac{1}{98.99.100}\)
2A = \(\frac{2}{1.2.3}\) + \(\frac{2}{2.3.4}\) + ... + \(\frac{2}{98.99.100}\)
2A = \(\frac12\).(\(\frac{2}{1.3}\)) + \(\frac13\).(\(\frac{2}{2.4}\)) + ... + \(\frac{1}{99}\).(\(\frac{2}{98.100}\))
2A = \(\frac12\).(\(\frac11-\frac13\)) + \(\frac13\).(\(\frac12-\frac14\)) + ...+ \(\frac{1}{99}\).(\(\frac{1}{98}-\frac{1}{100}\))
2A = \(\frac{1}{1.2}\) - \(\frac{1}{2.3}\) + \(\frac{1}{2.3}\) - \(\frac{1}{3.4}\) + ...+\(\frac{1}{98.99}\) - \(\frac{1}{99.100}\)
2A = \(\frac12-\frac{1}{9900}\)
2A = \(\frac{4949}{9900}\)
A = \(\frac{4949}{9900}\) : 2
A = 4949/19800
bài 1:
<=> \(x\left(x^2-\frac{9}{16}\right)=0\)
TH1:x=0
TH2: \(x^2-\frac{9}{16}=0\)
=> \(x^2=\frac{9}{16}\)
TH2a: \(\Rightarrow x=\frac34\)
\(TH2b:x=-\frac34\)
bài 2:
1) <=> \(2N=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\cdots+\frac{2}{98\cdot99\cdot100}\)
Áp dụng công thức: \(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+1\right)}\) ta có:
\(2N=\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)+\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+\cdots+\left(\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\)
\(2N=\frac{1}{1\cdot2}-\frac{1}{99\cdot100}\)
\(2N=\frac{4949}{9900}\)
\(\Rightarrow N=\frac{4949}{19800}\)
2) <=> \(5N=5^2+5^3+5^4+\cdots+5^{101}\)
=> \(5N-N=\left(5^2+5^3+5^4+\cdots+5^{101}\right)-\left(5+5^2+5^3+\cdots+5^{99}+5^{100}\right)\)
\(4N=5^{101}-5\)
=> \(N=\frac{\left(5^{101}-5\right)}{4}\)
Đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(A=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(A=\frac{1}{2}.\left(\frac{4950-1}{9900}\right)=\frac{1}{2}.\frac{4949}{9900}=\frac{4949}{19800}\)
Ủng hộ mk nha!!
2A=\(\frac{2}{1\cdot2\cdot3}\)+\(\frac{2}{2\cdot3\cdot4}\)+\(\frac{2}{3\cdot4\cdot5}\)+...+\(\frac{2}{2014\cdot2015\cdot2016}\)
2A=\(\frac{1}{1\cdot2}\)-\(\frac{1}{2\cdot3}\)+\(\frac{1}{2\cdot3}\)-\(\frac{1}{3\cdot4}\)+\(\frac{1}{3\cdot4}\)-\(\frac{1}{4\cdot5}\)+...+\(\frac{1}{2014\cdot2015}\)-\(\frac{1}{2015\cdot2016}\)
2A=\(\frac{1}{2}\)-\(\frac{1}{2015\cdot2016}\)
A=(\(\frac{1}{2}\)-\(\frac{1}{2015\cdot2016}\)):2
A=\(\frac{1}{2}\):2-\(\frac{1}{2015\cdot2016}\):2
A=\(\frac{1}{4}\)-\(\frac{1}{2015\cdot2016\cdot2}\)<\(\frac{1}{4}\)
Vậy A<\(\frac{1}{4}\)
A = \(\frac{2}{1.2.3}\) + \(\frac{2}{2.3.4}\) + ... + \(\frac{2}{98.99.100}\)
A = \(\frac12\).(\(\frac{2}{1.3}\)) + \(\frac13\).(\(\frac{2}{2.4}\)) + ... + \(\frac{1}{99}\).(\(\frac{2}{98.100}\))
A = \(\frac12\).(\(\frac11-\frac13\)) + \(\frac13\).(\(\frac12-\frac14\)) + ...+ \(\frac{1}{99}\).(\(\frac{1}{98}-\frac{1}{100}\))
A = \(\frac{1}{1.2}\) - \(\frac{1}{2.3}\) + \(\frac{1}{2.3}\) - \(\frac{1}{3.4}\) + ...+\(\frac{1}{98.99}\) - \(\frac{1}{99.100}\)
A = \(\frac12-\frac{1}{9900}\)
A = \(\frac{4949}{9900}\)
làm tiếp theo
\(S=\frac{5}{2}.\left(\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}+\frac{2}{99.100.101}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+....+\frac{1}{98.99}-\frac{1}{99.100}+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{2.3}-\frac{1}{100.101}\right)\)
còn lại tự làm
\(S=\frac{5}{2\cdot3\cdot4}+\frac{5}{3\cdot4\cdot5}+......+\frac{5}{99\cdot100\cdot101}\)
\(S\frac{2}{5}=\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+.....+\frac{2}{99\cdot100\cdot101}\)
\(\frac{2}{2\cdot3\cdot4}=\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\)
\(\frac{2}{3\cdot4\cdot5}=\frac{1}{3\cdot4}-\frac{1}{4\cdot5}\)
.............
\(\frac{2}{99\cdot100\cdot101}=\frac{1}{99\cdot100}-\frac{1}{100\cdot101}\)
\(\Rightarrow S\frac{2}{5}=\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+.........+\frac{1}{99\cdot100}-\frac{1}{100\cdot101}\)
\(\Rightarrow S\frac{2}{5}=\frac{1}{2\cdot3}-\frac{1}{100\cdot101}\)
\(\Rightarrow S\frac{2}{5}=\frac{1}{6}-\frac{1}{10100}\)
\(\Rightarrow S\frac{2}{5}=\frac{5047}{30300}\)
\(\Rightarrow S=\frac{5047}{30300}:\frac{2}{5}\)
\(\Rightarrow S=\frac{5047}{30300}\cdot\frac{5}{2}\)
\(\Rightarrow S=\frac{5047}{12120}\)