(19998*13-14443*18)*(1+2+.....+2008)
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Ta có: \(A=1-2+3-4+5-6+7-8+9\)
\(=(1+9)-(2+8)+(3+7)-(4+6)+5\)
\(=10-10+10-10+5\)
\(=5\)
Vậy \(A=5\)
B = 12 - 14 + 16 - 18 + ... + 2008 - 2010
B = -2 + (-2)+ (-2)+ (-2) + ...+ (-2)
B = -2 . 100
B = -200
\(\frac{x-1}{2013}+\frac{x-2}{2012}+\frac{x-3}{2011}=\frac{x-4}{2010}+\frac{x-5}{2009}+\frac{x-6}{2008}\) ( có lẽ đề như này )
\(\Leftrightarrow\frac{x-1}{2013}-1+\frac{x-2}{2012}-1+\frac{x-3}{2011}-1=\frac{x-4}{2010}-1+\frac{x-5}{2009}-1+\frac{x-6}{2008}-1\)
\(\Leftrightarrow\frac{x-2014}{2013}+\frac{x-2014}{2012}+\frac{x-2014}{2011}-\frac{x-2014}{2010}-\frac{x-2014}{2009}-\frac{x-2014}{2008}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
\(\Leftrightarrow x-2014=0\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\right)\)
\(\Leftrightarrow x=2014\)
...
Ta có : \(x^2+9x+20=x^2+4x+5x+20=\left(x+4\right)\left(x+5\right)\)
\(x^2+11x+30=x^2+5x+6x+30=\left(x+5\right)\left(x+6\right)\)
\(x^2+13x+42=x^2+6x+7x+42=\left(x+6\right)\left(x+7\right)\)
\(\Rightarrow Pt\Leftrightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\) (*)\(ĐKXĐ:x\ne-4;x\ne-5;x\ne-6;x\ne-7\)
(*) \(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow3.18=x^2+4x+7x+28\)
\(\Leftrightarrow x^2-2x+13x-26=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+13=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\x=-13\left(tm\right)\end{cases}}}\)
a, 2013/2018 < 2012/2018
b, 2013/2008 < 2008/2003
c,24/47 > 13/27
d,37/23 < 42/22
e 1/2 > 1/2017
g, 12/13 > 6/7
a) 3 + 8 + 13 + 18 + ... + 2008 + 2013
Áp dụng công thức tính dãy số ta có :
\(3+8+...+2013=\frac{\left[\left(2013-3\right):5+1\right].\left(2013+3\right)}{2}=\frac{403.2016}{2}=403.1008=406224\)
\(=\left(1+2+...+2008\right)\cdot\left(2\cdot3^2\cdot11\cdot13\cdot101-2\cdot3^2\cdot11\cdot13\cdot101\right)=0\)