I=\(\dfrac{2}{1x5}\)+\(\dfrac{2}{5x9}\)+\(\dfrac{2}{9x13}\)+.....+\(\dfrac{2}{181x185}\)
Những câu hỏi liên quan
\(\dfrac{4}{1x5}\) \(+\) \(\dfrac{4}{5x9}\) \(+\) \(\dfrac{4}{9x13}\) \(+\) ... \(+\) \(\dfrac{4}{53x57}\)
\(\dfrac{1}{1x5}\)+\(\dfrac{1}{5x9}\)+\(\dfrac{1}{9x13}\)+....+\(\dfrac{1}{45x49}\)
Tính nha
\(\dfrac{1}{1\times5}+\dfrac{1}{5\times9}+...+\dfrac{1}{45\times49}\)
\(=\dfrac{1}{4}\times\left(\dfrac{4}{1\times5}+\dfrac{4}{5\times9}+...+\dfrac{4}{45\times49}\right)\)
\(=\dfrac{1}{4}\times\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{45}-\dfrac{1}{49}\right)\)
\(=\dfrac{1}{4}\times\left(1-\dfrac{1}{49}\right)=\dfrac{1}{4}\times\dfrac{48}{49}=\dfrac{12}{49}\)
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Tính:
(2007 - 2005) + (2003 - 2001) +...+ (7 - 5) + (3 - 1)
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Viết tiếp 3 số hạng thứ 8;9 và 10 của dãy số sau: \(\dfrac{1}{1x5};\dfrac{1}{5x9};\dfrac{1}{9x13};...\)
s= \(\dfrac{1}{5x9}\)+\(\dfrac{1}{9x13}\)+....+\(\dfrac{1}{41x45}\)
\(S=\dfrac{1}{5\times9}+\dfrac{1}{9\times13}+...+\dfrac{1}{41\times45}\)
\(\Rightarrow4S=\dfrac{4}{5\times9}+\dfrac{4}{9\times13}+...+\dfrac{4}{41\times45}\)
\(\Rightarrow4S=\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}\)
\(\Rightarrow4S=\dfrac{1}{5}-\dfrac{1}{45}\)
\(\Rightarrow4S=\dfrac{8}{45}\)
\(\Rightarrow S=\dfrac{2}{45}\)
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\(=\dfrac{1}{4}\left(\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{8}{45}=\dfrac{2}{45}\)
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Adfrac{8}{1x5}+dfrac{8}{5x9}+dfrac{8}{9x13}+.....+dfrac{8}{25x29}
Bdfrac{3}{5x8}xdfrac{11}{8x9}+dfrac{12}{19x31}+dfrac{70}{31x101}+dfrac{99}{101x200}
Cdfrac{1}{8}+dfrac{1}{104}+dfrac{1}{234}+dfrac{1}{414}+dfrac{1}{644}+dfrac{1}{924}+dfrac{1}{1254}
Giúp mik với!! trước sáng mai nha, mik cần gấp lắm!! Cái này toán nâng cao nha, mik cảm ơn
Đọc tiếp
A=\(\dfrac{8}{1x5}+\dfrac{8}{5x9}+\dfrac{8}{9x13}+.....+\dfrac{8}{25x29}\)
B=\(\dfrac{3}{5x8}x\dfrac{11}{8x9}+\dfrac{12}{19x31}+\dfrac{70}{31x101}+\dfrac{99}{101x200}\)
C=\(\dfrac{1}{8}+\dfrac{1}{104}+\dfrac{1}{234}+\dfrac{1}{414}+\dfrac{1}{644}+\dfrac{1}{924}+\dfrac{1}{1254}\)
Giúp mik với!! trước sáng mai nha, mik cần gấp lắm!! Cái này toán nâng cao nha, mik cảm ơn
Tìm x:
\(\dfrac{7}{x}\)+\(\dfrac{4}{5x9}\)+\(\dfrac{4}{9x13}\)+.....+\(\dfrac{4}{41x45}\)=\(\dfrac{29}{45}\)
\(\Rightarrow\dfrac{7}{x}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{29}{45}\left(x\ne0\right)\\ \Rightarrow\dfrac{7}{x}+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{29}{45}\\ \Rightarrow\dfrac{7}{x}+\dfrac{8}{45}=\dfrac{29}{45}\\ \Rightarrow\dfrac{7}{x}=\dfrac{21}{45}\Rightarrow\dfrac{21}{3x}=\dfrac{21}{45}\Rightarrow3x=45\\ \Rightarrow x=15\)
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a) rút gọn: \(\dfrac{4^5x9^4-2x6^9}{2^{10}x3^8+6^8x20}\)
b) Cho A=\(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+\dfrac{4}{2^4}+\dfrac{5}{2^5}+...\dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}\).So sánh A với 2
a: \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
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tính 2/1x5+2/5x9+2/9x13+...+2/93x97+2/97x101 giúp mik với ak mik thi gấp lắm ạ mong mn giúp ạ cảm ơn mn rất rất nhìu ạ
`2/[1xx5]+2/[5xx9]+2/[9xx13]+....+2/[93xx97]+2/[97xx101]`
`=1/2xx(4/[1xx5]+4/[5xx9]+4/[9xx13]+....+4/[93xx97]+4/[97xx101])`
`=1/2xx(1-1/5+1/5-1/9+1/9-1/13+...+1/93-1/97+1/97-1/101)`
`=1/2xx(1-1/101)`
`=1/2xx100/101`
`=50/101`
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bn ghi rõ đc ko ạ mik ko hiểu lắm c.ơn bn
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1x5+5x9+9x13+...+81x85
giúp tui với
Lời giải:
$12A=1.5.12+5.9.(13-1)+9.13(17-5)+13.17(21-9)+....+77.81(85-73)+81.85(89-77)$
$=60+(5.9.13+9.13.17+13.17.21+...+77.81.85+81.85.89)-(1.5.9+5.9.13+9.13.17+...+73.77.81+77.81.85)$
$=60+81.85.89 - 1.5.9=612780$
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A = 1.5 + 5.9 + 9.13 + ... + 81.85
A = \(\dfrac{12}{12}\)(1.5 + 5.9 + 9.13 + 81.85)
A = \(\dfrac{1}{12}\).(1.5.12 + 5.9.12.+ 9.13.12 + ...+ 81.85.12]
A = \(\dfrac{1}{12}\).[1.5.(9 + 3) + 5.9.(13 - 1) + 9.13.(17 - 5) +...+ 81.85.(89 - 77)]
A = \(\dfrac{1}{12}\).[1.5.9 + 1.3.5 + 5.9.13 - 5.9.1 + 9.13.17 - 9.13.5 + ...+ 81.85.89 - 81.85.77]
A = \(\dfrac{1}{12}\).[1.3.5 + 81.85.89]
A = \(\dfrac{1}{12}\).[15 + 612765]
A = \(\dfrac{1}{12}\).612780
A = 51065
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