tính bằng cách hợp lí :
\(\frac{1}{2001\times2003}+\frac{1}{2003\times2005}+\frac{1}{2005\times2007}+.....+\frac{1}{2009\times2011}+\frac{1}{2011\times2013}\)
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mỗi số hạng trong biểu thức A đều nhỏ hơn 1 mà có 15 số nên tổng A sẽ nhỏ hơn 15
ta thay tong tren <1+1+1+1+1+1+1+1+1+1+1+1+1+1+1
hay tong tren be hon 15
Ta có: \(\frac{2004\cdot2007+6}{2005\cdot2005+2009}=\frac{\left(2005-1\right)\cdot2007+6}{2005\cdot2005+2009}=\frac{2005\cdot2007-1\cdot2007+6}{2005\cdot2005+2009}=\frac{2005\cdot2007-2007+6}{2005\cdot2005+2009}\)
\(=\frac{\text{2005 x (2005 + 2) - 2007 + 6}}{\text{2005 x 2005 + 2009}}=\frac{\text{2005 x 2005 + 2005 x 2 - 2007 + 6}}{\text{2005 x 2005 + 2009}}=\frac{\text{2005 x 2005 + 4010 - 2007 + 6}}{\text{2005 x 2005 + 2009}}=\text{ }\frac{\text{2005 x 2005 + 2009}}{\text{2005 x 2005 + 2009}}=1\)
minh lam duoc roi . cach viet phan so ban bam vao o mau vang o cuoi trang .cu di con chuot xuong cuoi trang thi thay 1 o vang , vao xem huong dan la biet ngay ma.
\(D=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}\right):\left(\frac{2011}{1}+\frac{2010}{2}+...+\frac{1}{2011}\right)\)
\(\Rightarrow D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
\(\Rightarrow D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)+1}\)
\(\Rightarrow D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}}\)
\(\Rightarrow D\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{2012\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)}\)
\(\Rightarrow D=\frac{1}{2012}\)
\(\frac{2005\cdot2004-1}{2003\cdot2005+2004}\)
\(=\frac{2005\cdot\left(2003+1\right)-1}{2003\cdot2005+2004}\)
\(=\frac{2005\cdot2003+2005-1}{2003\cdot2005+2004}\)
\(=\frac{2005\cdot2003+2004}{2003\cdot2005+2004}\)
\(=1\)
2005 x 2004 - 1 / 2003 × 2005 + 2004
= 2005 × (2003 + 1) - 1 / 2003 × 2005 + 2004
= 2005 × 2003 + (2005 - 1) / 2003 × 2005 + 2004
= 2005 × 2003 + 2004 / 2003 × 2005 + 2004
= 1
Ta làm đơn giản :
A = \(\frac{2005x2005+1}{2005x2005x2005-1}=\frac{1}{2005-1}=\frac{1}{2004}\)
B = \(\frac{2005+1}{2005x2005-1}\)=\(\frac{2006}{4020024}=\frac{1}{2004}\)
\(\frac{1}{2004}=\frac{1}{2004}\)
Nên A = B
Ta có :
\(\frac{1}{2009}+\frac{2}{2009}+....+\frac{2008}{2009}\)
\(=\frac{1+2+....+2008}{2009}\)
\(=\frac{2017036}{2009}=1004\)
Ta có ; \(\frac{1}{2009}+\frac{2}{2009}+\frac{3}{2009}+......+\frac{2008}{2009}\)
\(=\frac{1+2+3+......+2008}{2009}\)
\(=\frac{2017036}{2009}=1004\)
Giúp mình nha
Ta có: 1/ 2001 . 2003 = 1/2001 - 1/2003...
=> 1/2001 - 1/2003 + 1/2003 - 1/2005 + 1/2005 - 1/2007 + ... +1/2009 - 1/2011 +1/2011 - 1/2013
= 1/2001 - 1/2013
= 4/ 1342671