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a.\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2015}\right)=\frac{1}{2}.\frac{2}{3}...\frac{2014}{2015}=\frac{1.2.3...2014}{2.3...2015}=\frac{1}{2015}\)
b.\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{128}-\frac{1}{256}=1-\frac{1}{256}=\frac{255}{256}\)
c.\(\frac{5}{2}+\frac{5}{4}+\frac{5}{8}+...+\frac{5}{256}=5\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\right)=5.\frac{255}{256}=\frac{1275}{256}\)
d.14,35+(13,7-13,6).1=14,35+0,1.1=14,35+0,1=14,45
A = \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{2015}\right)\)
A = \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2014}{2015}\)
A = \(\frac{1}{2015}\)
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\cdot...\cdot\left(1-\frac{1}{2015}\right)=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{2014}{2015}=\frac{1\cdot2\cdot3\cdot...\cdot2014}{2\cdot3\cdot...\cdot2014\cdot2015}=\frac{1}{2015}\)
( 2013 x 2014 + 2014 x 2015 + 2015 x 2016) x ( 1 + 1/3 - 4/3)
=( 2013 x 2014 + 2014 x 2015 + 2015 x 2016) x ( 4/3 - 4/3)
=( 2013 x 2014 + 2014 x 2015 + 2015 x 2016) x 0
=0
Ta có: \(\left(2013\cdot2014+2014\cdot2015+2015\cdot2016\right)\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)\)
\(=\left(2013\cdot2014+2014\cdot2015+2015\cdot2016\right)\left(\dfrac{3}{3}+\dfrac{1}{3}-\dfrac{4}{3}\right)\)
=0
=1/2 x 2/3 x 3/4 x 4/5 x 5/6 x.....x2013/2014 - 2014/2015
=1/2014 - 2014/2015
=1/2015
a) 20,8 x 45 + 0,37 x 15 + 20,8 x 55 x 0,63
= 20,8 x ( 45 + 55 x 0,63 ) + 0,37 x 15
= 20,8 x ( 45 + 34,65 ) + 5,55
= 20,8 x 79,65 + 5,55
= 1656,72 + 5,55
= 1662,27
b) ( 2013 x 2014 + 2014 x 2015 + 2015 x 2016 ) x ( 1 + 1/3 - 1 và 1/3 )
= ( 2013 x 2014 + 2014 x 2015 + 2015 x 2016 ) x [( 1 - 1 ) + ( 1/3 - 1/3 ) ]
= ( 2013 x 2014 + 2014 x 2015 + 2015 x 2016 ) x 0
= 0
là lấy cái này cộng cái này rồi nhân cái này sẽ ra kết quả