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\(C=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{\frac{5}{2008}-\frac{5}{2009}-\frac{5}{2010}}+\frac{\frac{2}{2007}-\frac{2}{2008}-\frac{2}{2009}}{\frac{3}{2007}-\frac{3}{2008}-\frac{3}{2009}}\)
\(=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{5.\left(\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)}+\frac{2.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}{3.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}\)
\(=\frac{1}{5}+\frac{2}{3}\)
\(=\frac{13}{15}\)
Đặt S=\(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2008}}\)
5S=\(1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2007}}\)
5S-S=\(1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2007}}\)-\(\left(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2008}}\right)\)
4S=\(1-\frac{1}{5^{2008}}\)
=> S=\(\frac{1-\frac{1}{5^{2008}}}{4}\)
n) Theo bài ra ta có: \(\frac{x+1}{2008}=\frac{502}{x+1}\)
=> (x+1).(x+1) = 2008.502
=> (x+1)2 = 1008016
=> (x+1)2 = 10042
=> x+1 = 1004
=> x = 2004-1
=> x = 2003
Vậy x = 2003
p) Theo bà ra ta có: \(\left|\frac{5}{4}.x-\frac{7}{2}\right|-\left|\frac{5}{8}.x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}.x-\frac{7}{2}\right|=\left|\frac{5}{8}.x+\frac{3}{5}\right|\)
=> \(\frac{5}{4}.x-\frac{7}{2}=\pm\left(\frac{5}{8}.x+\frac{3}{5}\right)\)
=> \(\left[\begin{array}{nghiempt}\frac{5}{4}.x-\frac{7}{2}=\frac{5}{8}.x+\frac{3}{5}\\\frac{5}{4}.x-\frac{7}{2}=\frac{-5}{8}.x-\frac{3}{5}\end{array}\right.\)
=> \(\left[\begin{array}{nghiempt}\frac{5}{4}.x-\frac{5}{8}.x=\frac{3}{5}+\frac{7}{2}\\\frac{5}{4}.x+\frac{5}{8}.x=\frac{-3}{5}+\frac{7}{2}\end{array}\right.\)
=> \(\left[\begin{array}{nghiempt}\frac{5}{8}.x=\frac{41}{10}\\\frac{15}{8}.x=\frac{29}{10}\end{array}\right.\)
=> \(\left[\begin{array}{nghiempt}x=\frac{164}{25}\\x=\frac{116}{75}\end{array}\right.\)
Vậy x=\(\frac{164}{25}\) hoặc x=\(\frac{116}{75}\)
a) \(x-\frac{2}{5}=\frac{5}{7}\)
\(x=\frac{2}{5}+\frac{5}{7}\)
\(x=\frac{14}{35}+\frac{25}{35}=\frac{39}{35}\)
b)
\(\frac{-2}{5}x=\frac{4}{15}\)
\(x=\frac{4}{15}:-\frac{2}{5}\)
\(x=\frac{4}{15}\cdot-\frac{5}{2}=-\frac{2}{3}\)
c) \(2x\left(x-\frac{1}{7}\right)=2x^2-\frac{2x}{7}\)
d) \(\frac{1}{2}+\frac{3}{4}x=\frac{1}{4}\)
\(\frac{3}{4}x=\frac{1}{4}-\frac{1}{2}\)
\(\frac{3}{4}x=-\frac{1}{4}\)
\(x=-\frac{1}{4}\cdot\frac{4}{3}=-\frac{1}{3}\)
f) \(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{5}\)
\(\frac{2}{5}+x=\frac{11}{12}-\frac{2}{5}=\frac{31}{60}\)
\(x=\frac{31}{60}-\frac{2}{5}=\frac{7}{60}\)
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )
1
a) \(3\frac{1}{2}-\frac{1}{2}\cdot\left(-4,25-\frac{3}{4}\right)^2\) : \(\frac{5}{4}\)
= \(3\cdot25:\frac{5}{4}\)
= \(3\cdot\left(25:\frac{5}{4}\right)\)
=\(3\cdot20\)
=60
b)=\(\frac{3}{7}\cdot\left(1\frac{1}{2}+0,5-9\right)\)
=\(\frac{3}{7}\cdot\left(-7\right)\)
=\(-3\)
c) =
5G= 1+1/5+1/5^2+.....+1/5^2007
4G=5G-G=(1+1/5+1/5^2+....+1/5^2007)-(1/5+1/5^2+1/5^3+....+1/5^2008)
= 1 - 1/5^2008
=>G=(1-1/5^2008)/4
\(G=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2008}}\)(1)
\(\Rightarrow5G=1+\frac{1}{5}+...+\frac{1}{5^{2007}}\)(2)
Lấy (2) trừ đi (1) ta có :
\(4G=1-\frac{1}{5^{2008}}\)
\(\Rightarrow G=\frac{\left(1-\frac{1}{5^{2008}}\right)}{4}\)