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a)\(\frac{2016}{2017}< 1;\frac{2015}{2016}< 1\)
b)\(\frac{2017}{2016}>1;\frac{2016}{2015}>1\)
=> \(\frac{2016}{2017}\)và
\(\frac{2016}{2017}< 1;\frac{2016}{2015}< 1\)
\(\frac{2017}{2016}>1;\frac{2016}{2015}>1\)
=> \(\frac{2016}{2017}\)và \(\frac{2015}{2016}\)< \(\frac{2017}{2016}\)và \(\frac{2016}{2015}\)
Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
a)\(=\frac{2017}{2016}.\frac{3}{4}-\frac{1}{2016}.\frac{3}{4}\)
\(=\frac{3}{4}\left(\frac{2017}{2016}-\frac{1}{2016}\right)\)
\(=\frac{3}{4}.1\)
\(=\frac{3}{4}\)
b)\(=\frac{2015}{2016}\left(\frac{1}{2}+\frac{1}{3}-\frac{5}{6}\right)\)
\(=\frac{2015}{2016}.0\)
\(=0\)
Ta có A= 1/2015 + 2/2016 + 3/2017 + ... +2016/4030- 2016
A= 2015-2014/2015 + 2016-2014/2016 +...+4030-2014/4030-2016
A= 2015/2015-2014/2015+ 2016/2016-2014/2016 + ..... +4030/4030-2014/4030 -2016
A= 1-2014/2015 + 1-2014/2016 +....+1-2014/4030 -2016
A= (1+1+1+1+........+1) -(2014/2015+2014/2016+......+2014/4030) -2016
A=2016 - 2014.(1/2015+1/2016+....+1/4030) -2016
A= (2016 - 2016 ) - 2014. ( 1/2015+1/2016+.....+1/4030)
A=-2014.(1/2015+1/2016+....+1/4030)
mà B = 1/2015+1/2016+....+1/4030
nên A : B = -2014
các bn hãy ủng hộ mk nhé !!! Thanks everyone!!!