(2013*2014+2014*2015+2015*2016)*(1+1/3-1 1/3)
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\(A=\left[1+\left(-2\right)\right]+\left[3+\left(-4\right)\right]+....+\left[2013+\left(-2014\right)+2015\right]\)
\(A=\left(-1\right)+\left(-1\right)+....+\left(-1\right)+2015\left(\text{1007 số hạng }\left(-1\right)\right)=1008\)
Đặt \(\dfrac{1}{5}+\dfrac{2013}{2014}+\dfrac{2015}{2016}=B;\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}=C\)
\(A=\left(B+1\right)\cdot C-B\cdot\left(C+1\right)\)
\(=BC+C-BC-B\)
=C-B
\(=\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}-\dfrac{1}{5}-\dfrac{2013}{2014}-\dfrac{2015}{2016}=-\dfrac{1}{10}\)
( 2013 x 2014 +2014 x 2015 + 2015 x 2016 ) x ( 1 + 1/3 - 1 - 1/3 )
= ( 2013 x 2014 + 2014 x 2015 + 2015 x 2016 ) x 0
= 0
( 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
Tính
1 1/3 là hỗn số đúng không ta? :))
\(\left(2013\times2014+2014\times2015+2015\times2016\right)\times\left(1+\frac{1}{3}-1\frac{1}{3}\right)\)
\(=\left(2013\times2014+2014\times2015+2015\times2016\right)\times\left(1\frac{1}{3}-1\frac{1}{3}\right)\)
\(=\left(2013\times2014+2014\times2015+2015\times2016\right)\times0\)
\(=0\)