TínhB=3+3²+3³+...+3^100
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Sửa đề: 3/91*96
\(B=\dfrac{3}{1\cdot6}+\dfrac{3}{6\cdot11}+...+\dfrac{3}{91\cdot96}\)
=3/5(5/1*6+5/6*11+...+5/91*96)
=3/5(1-1/6+1/6-1/11+...+1/91-1/96)
=3/5*95/96=57/96=19/32
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\(B=1+\dfrac{1}{2}\cdot\dfrac{2\cdot3}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot4}{2}+...+\dfrac{1}{50}\cdot\dfrac{50\cdot51}{2}\)
\(=1+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{51}{2}\)
\(=\dfrac{50\cdot\dfrac{\left(51+2\right)}{2}}{2}=50\cdot\dfrac{53}{4}=662.5\)
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b) 2 - 8/9 = 10/9
c) 5/8 x 3 = 15/8
d) 5 : 6/21= 105/6
đ) 3/5 x 4/5 x 6/7= 72/175
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\(\dfrac{1}{x^2-x}+\dfrac{x-3}{x^2-1}\left(x\ne\pm1;x\ne0\right)\)
\(=\dfrac{1}{x\left(x-1\right)}+\dfrac{x-3}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x+1}{x\left(x-1\right)\left(x+1\right)}+\dfrac{x\left(x-3\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x+1+x^2-3x}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2-2x+1}{\left(x-1\right)\left[x\left(x+1\right)\right]}\)
\(=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x^2+x\right)}\)
\(=\dfrac{x-1}{x^2+x}\)
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\(B=\left(17+\dfrac{1}{25}-16-\dfrac{11}{15}\right)\cdot\dfrac{75}{23}+\dfrac{75}{23}\cdot\dfrac{23}{75}\)
\(=\dfrac{23}{75}\cdot\dfrac{75}{23}+1=1+1=2\)
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Đặt A = 1 + 2 + 22 + ... + 22008
2A = 2 + 22 + 23 + ... + 22009
2A - A = A = 1 - 22009
=> B = 1 - 22009 / 1 - 22009
=> B = 1
Tk mk nha!
Đặt C\(=1+2+2^2+....+2^{2008}\)
2C\(=2+2^2+2^3+....+2^{2009}\)
2C-C\(=2^{2009}-1\)
C\(=2^{2009}-1\)
Vậy B=\(\frac{2^{2009}-1}{1-2^{2009}}\)