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\(A=\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>\frac{2001}{2001}+\frac{2002}{2002}+\frac{2003}{2003}+\frac{2004}{2004}+\frac{2005}{2005}+\frac{2006}{2006}+\frac{2007}{2007}+\frac{2008}{2008}\)
\(A=\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>1+1+1+1+1+1+1+1\)\(A=\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>8\)
\(A>8\)
5/6 < 5/4
7/3 > 7/5
2001/2002 < 2001/2000
222/555 < 444/333
2003 / 2001 = 1 + 2/2001
1999/1997 = 1 + 2/1997
vì 2/ 2001 < 2/1997
nên 1 + 2/2001 < 1 + 2/1997
hay 2003 < 1999/1997
b, = 5/9 x 1/4 + 4/9 x 1/4
= 1/4 x ( 5/9 + 4/9 )
= 1/4 x 1
= 1/4
* Ý a mk k nhớ cách làm ^^, xl *
\(b,\dfrac{5}{9}\times\dfrac{1}{4}+\dfrac{4}{9}\times\dfrac{3}{12}\)
\(=\dfrac{5}{9}\times\dfrac{1}{4}+\dfrac{4}{9}\times\dfrac{1}{4}\)
\(=\dfrac{1}{4}\times\left(\dfrac{5}{9}+\dfrac{5}{9}\right)\)
\(=\dfrac{1}{4}\times\dfrac{9}{9}=\dfrac{1}{4}\times1=\dfrac{1}{4}\)
2000*2001*2002*2003*2004*2005*2006*2007*2008*2009*2010
=(2000+2010)*(2002+2008)*(2003+2007)*(2004+2006)*2005
=4010*4010*4010*4010*2005
=(4010+4010)*(4010+4010)*2005
=64320400*2005
=128962400200000000000
cách học sinh lớp 5
1 - 2000/2001 = 1/2001 ; 1 - 2001/2002 = 1/2002
2000/2001 < 2001/2002
Ta có :
\(A=\frac{3}{4.5}+\frac{3}{5.6}+\frac{3}{6.7}+...+\frac{3}{99.100}\)
\(A=3\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\right)\)
\(A=3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=3\left(\frac{1}{4}-\frac{1}{100}\right)\)
\(A=3.\frac{6}{25}\)
\(A=\frac{18}{25}\)
Vậy \(A=\frac{18}{25}\)
Chúc bạn học tốt ~
\(A=\frac{3}{4.5}+\frac{3}{5.6}+\frac{3}{6.7}+...+\frac{3}{99.100}\)
\(\Rightarrow A=3.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\right)\)
\(\Rightarrow A=3.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Rightarrow A=3.\left(\frac{1}{4}-\frac{1}{100}\right)=\frac{3.24}{100}\)
\(=\frac{3.4.6}{25.4}\)
\(\Rightarrow A=\frac{18}{25}\)