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= \(\frac{2.2}{1.3}+\frac{3.3}{2.4}+\frac{4.4}{3.5}+\frac{5.5}{4.6}+\frac{6.6}{5.7}\)
= \(\frac{2.3.4.5.6}{1.2.3.4.5}+\frac{2.3.4.5.6}{3.4.5.6.7}\)
= \(\frac{2}{1}+\frac{6}{7}\)
= 2\(\frac{6}{7}\)
Mình nghĩ zậy !!!!!!!!!!!!!!!!!!
Ta có: A = 1.2.3+3.4.5+5.6.7+...+99.100.101
A = 1.3 (5-3) + 3.5 (7-3) + 5.7 (9-3) + ............ + 99.101 (103 - 3)
A = (1.3.5 + 3.5.7 + 5.7.9 + .......... + 99.101.103) - (1.3.3 + 3.5.3 + ....... + 99.101.3)
A = (15+99.101.103.105) : 8 - 3.(1.3 + 3.5 +5.7 + ...... + 99.101)
A = 13517400 - 3.171650
A = 13002450
B)8*2*0,125*1/4*1/2*4
=(8*0,125)*(2*1/2)*(1/4*4)
=1*1*1
=1
\(A=\dfrac{1}{2}+\dfrac{3-2}{3.2}+\dfrac{4-3}{3.4}+...+\dfrac{100-99}{100.99}\)
\(A=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=1-\dfrac{1}{100}\)
\(A=\dfrac{99}{100}\)
\(2B=\dfrac{2}{1.3}+\dfrac{2}{3.5}+....+\dfrac{2}{2007.2009}+\dfrac{2}{2009..2011}\)
\(2B=\dfrac{3-1}{1.3}+\dfrac{5-3}{3,5}+...+\dfrac{2009-2007}{2009.2007}+\dfrac{2011-2009}{2011.2009}\)
\(2B=\dfrac{3}{3}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2007}-\dfrac{1}{2009}+\dfrac{1}{2009}-\dfrac{1}{2011}\)
\(2B=1-\dfrac{1}{2011}\)
\(2B=\dfrac{2010}{2011}\)
\(B=\dfrac{2010}{4022}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2009.2011}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{2009}-\frac{1}{2011}\)
\(=1-\frac{1}{2011}=\frac{2010}{2011}\)
a: \(=\dfrac{54-34}{189-119}=\dfrac{20}{70}=\dfrac{2}{7}\)
b: \(=\dfrac{6+6\cdot4+6\cdot49}{15+15\cdot4+15\cdot49}=\dfrac{6}{15}=\dfrac{2}{5}\)
c: \(=\dfrac{13\left(3-18\right)}{40\left(15-2\right)}=\dfrac{-15}{40}=-\dfrac{3}{8}\)
A=1/2{(1/1*2-1/2*3)+(1/2*3-1/3*4)+(1/3*4-1/4*5)+...+(1/18*19-1/19*20)}
=1/2{1/1*2-1/19*20}
=1/2*189/380
=189/760
vì 189/760<1/4
nên A=...<1/4
