\(A=\frac{3}{1^2x2^2}+\frac{5}{2^2x3^2}+\frac{7}{3^2x4^2}+............+\frac{19}{9^2x10^2}\)
Những câu hỏi liên quan
\(A=\frac{3}{1^2x2^2}+\frac{5}{2^2x3^2}+\frac{7}{3^2x4^2}+......+\frac{39}{19^2x20^2}\)
Tính A
\(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{39}{19^2.20^2}\)
\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{19^2}-\frac{1}{20^2}\)
\(=1-\frac{1}{20^2}\)
\(=\frac{399}{400}\)
Chứng minh \(\frac{3}{1^2x2^2}+\frac{5}{2^2x3^2}+\frac{7}{3^2x4^2}+...+\frac{2013}{1006^2x1007^2}\) < 1
tính:
a) \(\frac{1^2}{1x2}+\frac{2^2}{2x3}+\frac{3^2}{3x4}+...+\frac{100^2}{100x101}\)
b) \(\frac{2^2}{1x3}+\frac{3^2}{2x4}+\frac{4^2}{3x5}+...+\frac{59^2}{58x60}\)
a) Đặt \(A=\frac{1^2}{1.2}+\frac{2^2}{2.3}+.........+\frac{100^2}{100.101}\)
\(\Rightarrow A=\left(1^2+2^2+..........+100^2\right)\)\(.\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{100.101}\right)\)
\(\Rightarrow A=\left(1^2+2^2+......+100^2\right).\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow A=\left(1^2+2^2+......+100^2\right).\left(1-\frac{1}{101}\right)\)
\(\Rightarrow A=\left(1^2+2^2+.....+100^2\right).\left(\frac{100}{101}\right)\)(a)
Đặt \(M=\left(1^2+2^2+........+100^2\right)\)
\(\Rightarrow M=1.1+2.2+.....+100.100\)
\(\Rightarrow M=1.\left(2-1\right)+2.\left(3-1\right)+....+100.\left(101-1\right)\)
\(\Rightarrow M=\left(1.2-1\right)+\left(2.3-2\right)+.....+\left(100.101-100\right)\)
\(\Rightarrow M=\left(1.2+2.3+.....+100.101\right)-\left(1+2+......+100\right)\)
\(\Rightarrow M=\left(1.2+2.3+......+100.101\right)-5050\)(1)
Đặt \(N=1.2+2.3+....+100.101\)
\(\Rightarrow3.N=1.2.3+2.3.3+......+100.101.3\)
\(\Rightarrow3N=1.2.\left(3-0\right)+2.3.\left(4-1\right)+......+100.101.\left(102-99\right)\)
\(\Rightarrow3N=\left(1.2.3-0\right)+\left(1.2.3-2.3.4\right)+.......+\left(100.101.102-100.101.99\right)\)
\(\Rightarrow3N=100.101.102-0\)
\(\Rightarrow N=343400\)
Thay N = 343400 vào 1) ta được:
M = 343400 - 5050
=> M = 338350
Thay M = 338350 Vào (a) ta được:
A = 338350 . \(\frac{100}{101}\)
=> \(A=\frac{33835000}{101}\)
Vậy \(\frac{1^2}{1.2}+\frac{2^2}{2.3}+.........+\frac{100^2}{100.101}=\frac{33835000}{101}=335000\)
b) Đặt \(B=\frac{2^2}{1.3}+\frac{3^2}{2.4}+..........+\frac{59^2}{58.60}\)
\(\Rightarrow B=\left(2^2+3^2+........+59^2\right).\left(\frac{1}{1.3}+\frac{1}{2.4}+.....+\frac{1}{58.60}\right)\)
Đặt \(G=2^2+3^2+.........+59^2\)VÀ \(H=\frac{1}{1.3}+\frac{1}{2.4}+.........+\frac{1}{58.60}\)
\(\Rightarrow G=2.2+3.3+.......+59.59\) VÀ \(2.H=\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{58.60}\)
Rồi bạn làm như ở phần a) ý
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Rút gọn
1.\(\frac{7}{19}.\frac{8}{11}+\frac{7}{19}.\frac{3}{11}+\frac{12}{19}\)
2.\(\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{9}-\frac{4}{11}}\)
\(1\)) \(\frac{7}{19}.\frac{8}{11}+\frac{7}{19}.\frac{3}{11}+\frac{12}{19}\)
\(=\frac{7}{19}\left(\frac{8}{11}+\frac{3}{11}\right)+\frac{12}{19}\)
\(=\frac{7}{19}+\frac{12}{19}\)
\(=1\)
\(2\)) \(\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{9}-\frac{4}{11}}\)
\(=\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}}{2\left(\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}\right)}\)
\(=\frac{1}{2}\)
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1. \(\frac{7}{19}\times\frac{8}{11}+\frac{7}{19}\times\frac{3}{11}+\frac{12}{19}\)
\(=\frac{7}{19}\left(\frac{8}{11}+\frac{3}{11}\right)+\frac{12}{19}\)
\(=\frac{7}{19}\times1+\frac{12}{19}=\frac{19}{19}=1\)
2. \(\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{9}-\frac{4}{11}}\)
\(=\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}}{2\left(\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}\right)}=\frac{1}{2}\)
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1) Tính:a) frac{left(1+frac{17}{1}right).left(1+frac{17}{2}right).left(1+frac{17}{3}right).....left(1+frac{17}{19}right)}{left(1+frac{19}{1}right).left(1+frac{19}{2}right).left(1+frac{19}{3}right).....left(1+frac{19}{17}right)}b) frac{frac{-6}{5}+frac{6}{19}-frac{6}{23}}{frac{9}{5}-frac{9}{19}+frac{9}{23}}c) frac{frac{1}{6}-frac{1}{39}+frac{1}{51}}{frac{1}{8}-frac{1}{52}+frac{1}{68}}d) frac{frac{2}{3}-frac{2}{5}-frac{2}{7}+frac{2}{11}}{frac{13}{3}-frac{13}{5}-frac{13}{7}+frac{13}{11}}e) frac{fra...
Đọc tiếp
1) Tính:
a) \(\frac{\left(1+\frac{17}{1}\right).\left(1+\frac{17}{2}\right).\left(1+\frac{17}{3}\right).....\left(1+\frac{17}{19}\right)}{\left(1+\frac{19}{1}\right).\left(1+\frac{19}{2}\right).\left(1+\frac{19}{3}\right).....\left(1+\frac{19}{17}\right)}\)
b) \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}\)
c) \(\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}\)
d) \(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}+\frac{13}{11}}\)
e) \(\frac{\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2017}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}}\)
2) CMR: \(\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{800}}< \frac{1}{3}\)
b) \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}=\frac{\left(-6\right).\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}=\frac{-6}{9}=\frac{-2}{3}\)
d) \(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}+\frac{13}{11}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}=\frac{2}{13}\)
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Tính:a) frac{left(1+frac{17}{1}right).left(1+frac{17}{2}right).left(1+frac{17}{3}right).....left(1+frac{17}{19}right)}{left(1+frac{19}{1}right).left(1+frac{19}{2}right).left(1+frac{19}{3}right).....left(1+frac{19}{17}right)}b) frac{frac{-6}{5}+frac{6}{19}-frac{6}{23}}{frac{9}{5}-frac{9}{19}+frac{9}{23}}c) frac{frac{1}{6}-frac{1}{39}+frac{1}{51}}{frac{1}{8}-frac{1}{52}+frac{1}{68}}d) frac{frac{2}{3}-frac{2}{5}-frac{2}{7}+frac{2}{11}}{frac{13}{3}-frac{13}{5}-frac{13}{7}+frac{13}{11}}e) frac{frac{1...
Đọc tiếp
Tính:
a) \(\frac{\left(1+\frac{17}{1}\right).\left(1+\frac{17}{2}\right).\left(1+\frac{17}{3}\right).....\left(1+\frac{17}{19}\right)}{\left(1+\frac{19}{1}\right).\left(1+\frac{19}{2}\right).\left(1+\frac{19}{3}\right).....\left(1+\frac{19}{17}\right)}\)
b) \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}\)
c) \(\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}\)
d) \(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}+\frac{13}{11}}\)
e) \(\frac{\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2017}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}}\)
2) CMR: \(\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{800}}< \frac{1}{3}\)
Làm tiếp:
\(=\left(1+\frac{1}{2}+.....+\frac{1}{2017}\right)-\left(1+\frac{1}{2}+....+\frac{1}{1008}\right)\)
\(=\frac{1}{1009}+\frac{1}{1010}+.........+\frac{1}{2017}\)
\(\Rightarrow\frac{\frac{1}{1009}+....+\frac{1}{2017}}{1-\frac{1}{2}+.....+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}}=1\)
Bài 2:
Đặt \(A=\frac{1}{2^2}+.......+\frac{1}{2^{800}}\)
\(4A=1+\frac{1}{2^2}+.....+\frac{1}{2^{798}}\)
\(\Rightarrow4A-A=1-\frac{1}{2^{800}}\)
\(\Rightarrow3A=1-\frac{1}{2^{800}}< 1\Rightarrow A< \frac{1}{3}\)
Vậy \(\frac{1}{2^2}+\frac{1}{2^4}+........+\frac{1}{2^{800}}< \frac{1}{3}\)
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Bài 1:Tính
a, Xét biểu thức \(\frac{\left(1+\frac{n}{1}\right)\left(1+\frac{n}{2}\right).........\left(1+\frac{n}{n+2}\right)}{\left(1+\frac{n+2}{1}\right)\left(1+\frac{n+2}{2}\right)..........\left(1+\frac{n+2}{n}\right)}\) với\(n\in N\)
Ta có:\(\frac{\left(1+\frac{n}{1}\right)\left(1+\frac{n}{2}\right).......\left(1+\frac{n}{n+2}\right)}{\left(1+\frac{n+2}{1}\right)\left(1+\frac{n+2}{2}\right)......\left(1+\frac{n+2}{n}\right)}\)
\(=\frac{\frac{n+1}{1}.\frac{n+2}{2}........\frac{2n+2}{n+2}}{\frac{n+3}{1}.\frac{n+4}{2}.........\frac{2n+2}{n}}\)
\(=\frac{\frac{\left(n+1\right)\left(n+2\right).......\left(2n+2\right)}{1.2.3.........\left(n+2\right)}}{\frac{\left(n+3\right)\left(n+4\right)........\left(2n+2\right)}{1.2.3.........n}}\)
\(=\frac{\left(n+1\right)\left(n+2\right).......\left(2n+2\right).1.2.3.......n}{\left(n+3\right)\left(n+4\right)........\left(2n+2\right).1.2.3......\left(n+2\right)}\)
\(=\frac{\left(n+1\right)\left(n+2\right)}{\left(n+1\right)\left(n+2\right)}=1\)
Áp dụng vào bài toán ta có đáp số là:1
b, \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}=\frac{\left(-6\right).\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}=\frac{-6}{9}=-\frac{2}{3}\)
c,\(\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}=\frac{\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}{\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}=\frac{\frac{1}{3}}{\frac{1}{4}}=12\)
d,\(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}\right)}=\frac{2}{13}\)
e,Xét mẫu số ta có:
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..........+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\)
\(=1+\frac{1}{2}-2.\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-2.\frac{1}{4}+.....+\frac{1}{2015}+\frac{1}{2016}-2.\frac{1}{2016}+\frac{1}{2017}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{2017}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+.........+\frac{1}{2016}\right)\)
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\(A=\left(6:\frac{3}{5}-1\frac{1}{6}x\frac{6}{7}\right):\left(4\frac{1}{5}x\frac{10}{11}+5\frac{2}{11}\right)\)\(B=\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{4}\right)x.......x\left(1-\frac{1}{2015}\right)x\left(1-\frac{1}{2016}\right)\)
\(C=5\frac{9}{10}:\frac{3}{2}-\left(2\frac{1}{3}x4\frac{1}{2}-2x2\frac{1}{3}\right):\frac{7}{4}\)
Tính:a) Afrac{(1+17)(1+frac{17}{2})(1+frac{17}{3})...(1+frac{17}{19})}{(1+19)(1+frac{19}{2})(1+frac{19}{3})...(1+frac{19}{17})}b) Bfrac{1}{-2}.frac{1}{3}+frac{1}{-3}.frac{1}{4}+...+frac{1}{-5}.frac{1}{10}c) C(1-frac{1}{1.2})+(1-frac{1}{2.3})+...+(1-frac{1}{2015.2016})d) Dfrac{frac{9}{1}+frac{8}{2}+frac{7}{3}+...+frac{1}{9}}{frac{1}{2}+frac{1}{3}+frac{1}{3}+...+frac{1}{10}}
Đọc tiếp
Tính:
a) \(A=\frac{(1+17)(1+\frac{17}{2})(1+\frac{17}{3})...(1+\frac{17}{19})}{(1+19)(1+\frac{19}{2})(1+\frac{19}{3})...(1+\frac{19}{17})}\)
b) \(B=\frac{1}{-2}.\frac{1}{3}+\frac{1}{-3}.\frac{1}{4}+...+\frac{1}{-5}.\frac{1}{10}\)
c) \(C=(1-\frac{1}{1.2})+(1-\frac{1}{2.3})+...+(1-\frac{1}{2015.2016})\)
d) \(D=\frac{\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{1}{9}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{3}+...+\frac{1}{10}}\)
"ai giúp mình bài tính tổng này với .Trả lời rõ ra nhé : \(1\frac{2}{3}+4\frac{3}{5};\)
\(2\frac{5}{12}+3\frac{4}{9};\frac{7}{31}+\frac{14}{29}+\frac{27}{31};8\frac{3}{19}+7\frac{12}{19}+1\frac{4}{19};12\frac{3}{5}+\frac{17}{24}+\frac{7}{5};4\frac{2}{3}+2\frac{1}{5}+2\frac{2}{3}+3\frac{3}{5}\)