\(\frac{1+< 1+2>+< 1+2+3>+..............+< 1+2+3+4+...........+98>}{1\times98+2\times97+3\times96+..............................+98\times1}\)
Giúp mình nhé ai nhanh nhất mình tíck
Những câu hỏi liên quan
Cho B = \(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...........+\left(1+2+3+......+98\right)}{1\times98+2\times97+3\times96+.....+98\times1}\)
Rút gọn B
có tử bằng 1+(1+2)+(1+2+3)+...+(1+2+3+...+98)
vậy sẽ có 98 lần số 1 97 lần số 2 96 lần số 3 ... và 1 lần số 98
=> Tử bằng 1x98 + 2x97 + ... + 98x1 = mẫu
=> B=1
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a,A=\(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)
b,B=\(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{998\times999\times100}\)
c,C=\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+98\right)}{1\times98+2\times97+3\times96+...+98\times1}\)
Tính nhanh:
\(\frac{1}{99}-\frac{1}{99\times98}-\frac{1}{98\times97}-\frac{1}{97\times96}-...-\frac{1}{3\times2}\)
Các bạn vui lòng giải đầy đủ giúp mình. Thanks trước!
\(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-....-\frac{1}{3.2}\)
=\(\frac{1}{99}-\left(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{98.99}\right)\)
=\(\frac{1}{99}-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\right)\)
=\(\frac{1}{99}-\left(\frac{1}{2}-\frac{1}{99}\right)\)
=\(\frac{1}{99}-\frac{97}{198}\)
=\(\frac{-95}{198}\)
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Tính \(\dfrac{1}{100\times99}-\dfrac{1}{99\times98}-\dfrac{1}{98\times97}-...-\dfrac{1}{3\times2}-\dfrac{1}{2\times1}\)
\(\dfrac{1}{100.99}-\dfrac{1}{99.98}-...-\dfrac{1}{2.1}\)
\(=\dfrac{1}{99}-\dfrac{1}{100}-\dfrac{1}{98}+\dfrac{1}{99}-\dfrac{1}{97}+\dfrac{1}{98}-...-\dfrac{1}{2}+\dfrac{1}{3}-1+\dfrac{1}{2}\)
\(=\dfrac{2}{99}-\dfrac{1}{100}-1=-\dfrac{9799}{9900}\)
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Tính nhanh
C = \(\dfrac{1}{100}\) - \(\dfrac{1}{100\times99}\) - \(\dfrac{1}{99\times98}\) - \(\dfrac{1}{98\times97}\) - ... - \(\dfrac{1}{3\times2}\) - \(\dfrac{1}{2\times1}\)
C= 1/100-(1/1.2+1/2.3+...+1/97.98+1/98.99+1/99.100)
C=1/100-(1-1/2+1/2-1/3+...+1/97-1/98+1/98-1/99+1/99-1/100)
C=1/100-(1-1/100)
C=1/100-99/100
C=-98/100=-49/50
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\(C=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(=-\left(\dfrac{1}{100.99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)+\dfrac{1}{100}\)
\(=-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)+\dfrac{1}{100}\)
\(=-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)+\dfrac{1}{100}\)
\(=-\left(1-\dfrac{1}{100}\right)+\dfrac{1}{100}\)
\(=\left(-1\right)+\dfrac{1}{50}=-\dfrac{49}{50}\)
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theo bài ra ta có:
\(C=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\\ \Rightarrow C=\dfrac{1}{100}-\left(\dfrac{1}{100.99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\\ \Rightarrow C=\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\\ \Rightarrow C=\dfrac{1}{100}-\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\) \(\Rightarrow C=\dfrac{1}{100}-\left(\dfrac{1}{1}-\dfrac{1}{100}\right)\\ \Rightarrow C=\dfrac{1}{100}-\dfrac{99}{100}\\ \Rightarrow C=\dfrac{-98}{100}=\dfrac{-49}{50}\)
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Xem thêm câu trả lời
Tính:
A = \(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+99\right)}{1\times99+2\times98+3\times97+...+99\times1}\)
B = \(\frac{1\times2010+2\times2009+3\times2008+...+2010\times1}{\left(1+2+3+...+2010\right)+\left(1+2+3+...+2009\right)+...+\left(1+2\right)+1}\)
TÍnh A = 1+(1+2)+(1+2+3)+...+(1+2+3+...+98) / 1 x 98 + 2 x 97+ 3 x 96 + ...+ 98 x x1
giúp mình nhanh nhé mình tick cho mình đang cần gấp
D=(1*98+2*97+3*96+..+98*1)/(1*2+2*3+3*4+...98*99)
Giúp mình nhé!
tính nhanh
C = 1/100 - 1/100*99 - 1/99*98 -1/98*97 -........ -1/3*2 - 1/2*1
GIÚP MÌNH VỚI SONG KẾT BẠN VỚI MÌNH NHÉ!!!!!!!!
\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(C=\frac{1}{100}-\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{99-98}{98.99}+\frac{100-99}{99.100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{2}{100}-1=-\frac{49}{50}\)
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