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\(C=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{\frac{5}{2008}-\frac{5}{2009}-\frac{5}{2010}}+\frac{\frac{2}{2007}-\frac{2}{2008}-\frac{2}{2009}}{\frac{3}{2007}-\frac{3}{2008}-\frac{3}{2009}}\)
\(=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{5.\left(\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)}+\frac{2.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}{3.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}\)
\(=\frac{1}{5}+\frac{2}{3}\)
\(=\frac{13}{15}\)
a: \(=\dfrac{2008}{2007}-2009\cdot2-\dfrac{2009}{2007}+2009\cdot2\)
=-1/2007
b: \(=\dfrac{5^5\cdot5^3\cdot2^6-5^4\cdot5^3\cdot2^6+5^7\cdot2^{10}}{5^6\cdot2^{10}}\)
\(=\dfrac{5^8\cdot2^6-5^7\cdot2^6+5^7\cdot2^{10}}{5^6\cdot2^{10}}\)
\(=\dfrac{5^7\cdot2^6\left(5-1+2^4\right)}{5^6\cdot2^{10}}=\dfrac{5}{16}\cdot\dfrac{20}{1}=\dfrac{100}{16}=\dfrac{25}{4}\)
1. 2008.\(\left(\dfrac{1}{2007}-\dfrac{2009}{1004}\right)-2009\left(\dfrac{1}{2007}-2\right)\)
=\(\left(2008.\dfrac{1}{2007}-2008.\dfrac{2009}{1004}\right)-\left(2009.\dfrac{1}{2007}-2009.2\right)\)
=\(\left(\dfrac{2008}{2007}-2.2009\right)-\left(\dfrac{2009}{2007}-2.2009\right)\)
=\(\left(\dfrac{2008}{2007}-4018\right)-\left(\dfrac{2009}{2007}-4018\right)\)
=\(\dfrac{2008}{2007}-4018-\dfrac{2009}{2007}+4018\)
=\(\left(\dfrac{2008}{2007}-\dfrac{2009}{2007}\right)+\left[\left(-4018\right)+4018\right]\)
=\(\dfrac{1}{2007}.\left(2008-2009\right)+0\)
=\(\dfrac{1}{2007}.\left(-1\right)+0\)
=\(\dfrac{-1}{2007}\)
2.\(\dfrac{5^5.20^3-5^4.20^3+5^7.4^5}{\left(20+5\right)^3+4^5}\)
=\(\dfrac{5^5.\left(2^2.5\right)^3-5^4.\left(2^2.5\right)^3+5^7.\left(2^2\right)^5}{\left[\left(2^2.5\right)+5\right]^3+\left(2^2\right)^5}\)
=\(\dfrac{5^5.2^6.5^3-5^4.2^6.5^3+5^7.2^{10}}{2^6.5^3+5^3+2^{10}}\)
=\(\dfrac{5^9.2^6-5^7.2^6+5^7.2^{10}}{5^3.\left(2^6+1\right)+2^{10}}\)
=\(\dfrac{5^7.2^6\left(5^2-1-2^4\right)}{5^3\left(2^6+1\right)+2^{10}}\)
bí rồi
SỐ SỐ HẠNG LÀ:(2009-1):(3-1)+1=1005 TỔNG LÀ :(2009+1).1005:2