\(\dfrac{2}{1\cdot3}\)+\(\dfrac{2}{3\cdot5}\) +\(\dfrac{2}{5\cdot7}\)+\(\dfrac{2}{7\cdot9}\)+\(\dfrac{2}{9\cdot11}\) dấu . là nhân nh.mọi ng cho m lời giải chi tiết vs ạ.
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Hãy tính các tổng sau:
a)dfrac{1}{1cdot3}+dfrac{1}{3cdot5}+dfrac{1}{5cdot7}+dfrac{1}{7cdot9}+dfrac{1}{9cdot11}
b)dfrac{1}{4cdot7}+dfrac{1}{7cdot10}+dfrac{1}{10cdot13}+dfrac{1}{13cdot16}
c)dfrac{1}{2cdot7}+dfrac{1}{7cdot12}+dfrac{1}{12cdot17}+...
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Hãy tính các tổng sau:
a)\(\dfrac{1}{1\cdot3}\)+\(\dfrac{1}{3\cdot5}\)+\(\dfrac{1}{5\cdot7}\)+\(\dfrac{1}{7\cdot9}\)+\(\dfrac{1}{9\cdot11}\)=
b)\(\dfrac{1}{4\cdot7}\)+\(\dfrac{1}{7\cdot10}\)+\(\dfrac{1}{10\cdot13}\)+\(\dfrac{1}{13\cdot16}\)=
c)\(\dfrac{1}{2\cdot7}\)+\(\dfrac{1}{7\cdot12}\)+\(\dfrac{1}{12\cdot17}\)+...=
1100444-88888=
a)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(=\frac{1}{2}.\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+\frac{11-9}{9.11}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{11}\right)\)
\(\frac{10}{22}\)
1)A=\(\dfrac{5}{1\cdot2}+\dfrac{5}{2\cdot3}+.....+\dfrac{5}{99\cdot100}\)
C=\(1\cdot2\cdot3+2\cdot3\cdot4++3\cdot4\cdot5+4\cdot5\cdot6+5\cdot6\cdot7+6\cdot7\cdot8+7\cdot8\cdot9+8\cdot9\cdot10\)
D=\(1^2+2^2+3^2+...+99^2+100^2\)
a, A= \(5\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)
\(A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(A=5\left(1-\dfrac{1}{100}\right)\)
\(A=5.\dfrac{99}{100}=\dfrac{99}{20}.\)
b, \(C=1.2.3+2.3.4+...+8.9.10\)
\(4C=1.2.3.4+2.3.4.\left(5-1\right)+...+8.9.10.\left(11-7\right)\)\(4C=1.2.3.4+2.3.4.5-1.2.3.4+...+8.9.10.11-7.8.9.10\)\(4C=8.9.10.11\)
\(C=\dfrac{8.9.10.11}{4}=1980.\)
c, https://hoc24.vn/hoi-dap/question/384591.html
Câu này bạn vào đây mình đã giải câu tương tự nhé.
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\(1)A=\dfrac{5}{1.2}+\dfrac{5}{2.3}+...+\dfrac{5}{99.100}\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\cdot\dfrac{99}{100}\)
\(\Leftrightarrow A=\dfrac{99}{20}\)
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Tính:
\(\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}+\dfrac{1}{9\cdot11}+\dfrac{1}{11\cdot13}+\dfrac{1}{13\cdot15}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{13\cdot15}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{4}{15}=\dfrac{2}{15}\)
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\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{1}{15}=\dfrac{5}{15}-\dfrac{1}{15}=\dfrac{4}{15}\)
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= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{13}\)
= \(\dfrac{1}{3}-\dfrac{1}{13}\)
=\(\dfrac{10}{26}=\dfrac{5}{13}\)
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23 tháng 8 2023 lúc 21:06
\(B=\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+\dfrac{2}{3\cdot4\cdot5}+\dfrac{2}{4\cdot5\cdot6}+\dfrac{2}{5\cdot6\cdot7}+\dfrac{2}{6\cdot7\cdot8}\)
\(B=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+\dfrac{2}{4.5.6}+\dfrac{2}{5.6.7}+\dfrac{2}{6.7.8}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{6.7}-\dfrac{1}{7.8}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{7.8}\)
\(=\dfrac{1}{2}-\dfrac{1}{56}=\dfrac{27}{56}\)
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\(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{99\cdot101}\)
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.101}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(=\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)
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\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+......+\dfrac{2}{99.101}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+.....+\dfrac{1}{99}-\dfrac{1}{101}\)
\(=\dfrac{1}{3}-\dfrac{1}{101}\)
\(=\dfrac{101}{303}-\dfrac{3}{303}\)
\(=\dfrac{98}{303}\)
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Rút gọn biểu thức:
\(A=-\dfrac{2}{1\cdot3}-\dfrac{2}{3\cdot5}-\dfrac{2}{5\cdot7}-...-\dfrac{2}{19\cdot21}-\dfrac{2}{23\cdot25}-\dfrac{2}{25\cdot27}-\dfrac{1}{27}\)
ta được A = ...
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\(A=-B\)
\(B=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{23.25}+\dfrac{2}{25.27}+\dfrac{1}{27}\)
\(B=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{23}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{27}+\dfrac{1}{27}\)
\(B=1\)
A=-1
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\(A=-\dfrac{2}{1.3}-\dfrac{2}{3.5}-......-\dfrac{2}{25.27}-\dfrac{1}{27}\)
\(\Leftrightarrow A=-\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+.....+\dfrac{1}{27}\right)\)
\(\Leftrightarrow A=-\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{25}-\dfrac{1}{27}+\dfrac{1}{27}\right)\)
\(\Leftrightarrow A=-1\)
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= -(1-1/3+1/3-1/5+1/5-1/7+..+1/19-1/21+1/21-1/23+1/23-1/25+1/25-1/27+1/27)
= -1
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Tính nhanh \(\dfrac{8}{1\cdot3}-\dfrac{16}{3.5}+\dfrac{24}{5\cdot7}-\dfrac{32}{7\cdot9}+\dfrac{40}{9\cdot11}-...+\dfrac{384}{95\cdot97}-\dfrac{392}{97\cdot99}\)
Tính nhanh
\(\dfrac{8}{1\cdot3}-\dfrac{16}{3.5}+\dfrac{24}{5\cdot7}-\dfrac{32}{7\cdot9}+\dfrac{40}{9\cdot11}-...+\dfrac{384}{95\cdot97}-\dfrac{392}{97\cdot99}\)
Tính giá trị của các biểu thức sau 1) A1+2+2^2+...+2^{2015} 2) Bleft(dfrac{1}{4}-1right)cdotleft(dfrac{1}{9}-1right)cdotleft(dfrac{1}{16}-1right)cdotcdotcdotcdotcdotleft(dfrac{1}{400}-1right) 3) Cleft(dfrac{1}{4cdot9}+dfrac{1}{9cdot14}+dfrac{1}{14cdot19}+...+dfrac{1}{44cdot49}right)cdotdfrac{1-3-5-7-...-49}{89} 4) Ddfrac{2^{12}cdot3^5-4^6cdot9^2}{left(2^2cdot3right)^6+8^4cdot3^5}-dfrac{5^{10}cdot7^3-25^5cdot49^2}{left(125cdot7right)^3+5^9cdot14^3} 5) Edfrac{dfrac{1}{2003}+dfrac{1}{2004}-dfrac{1}...
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Tính giá trị của các biểu thức sau 1) \(A=1+2+2^2+...+2^{2015}\) 2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\) 3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\) 4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\) 5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\) 6) Cho 13+23+...+103=3025 Tính S= 23+43+63+...+203