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A = 1.2+2.3+...+2016.2017
3A=1.2.3 + 2.3.(4-1) + .. + 2016.2017.(2018-2015)
3A = 1.2.3 + 2.3.4 - 1.2.3 + ... + 2016.2017.2018 - 2015.2016.2017
3A = 2016.2017.2018
A = 2016.2017.2018 : 3
A = 2735245632
3A=1*2*3+2*3*(4-1)+.........+2016*2017.(2018-2015)
3A=1.2.3-1.2.3+2.3.4-2.3.4+.........+2016.2017.3
3A=2016.2017.2018
KẾT QUẢ TỰ TÍNH
A = \(\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+...+\dfrac{1}{2013.2014.2015.2016}\)
3A =\(\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+...+\dfrac{3}{2013.2014.2015.2016}\)
3A = \(\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+...+\dfrac{1}{2013.2014.2015}-\dfrac{1}{2014.2015.2016}\)
3A = \(\dfrac{1}{1.2.3}-\dfrac{1}{2014.2015.2016}\)
3A = \(\dfrac{2014.2015.2016-6}{6.2014.2015.2016}\)
A=\(\dfrac{2014.2015.2016-6}{6.2014.2015.2016}:3\)
A=\(\dfrac{2014.2015.2016-6}{6.2014.2015.2016}.\dfrac{1}{3}\)
A=\(\dfrac{2014.2015.2016-6}{9.2014.2015.2016}\)
Mình không muốn rút gọn hơn vì nó sẽ quá cồng kềnh nên mình để tạm thế này nhé!
bn xem có giúp gì được ko nha!
http://toantieuhoclh.violet.vn/entry/show/entry_id/10236071
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: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
toán lơp 6 nha mấy bạn giúp mình với
1.2+2.3+3.4+4.5+...+2015.2016
=>3(1.2+2.3+3.4+4.5+...+2015.2016)
=>1.2.3+2.3.3+3.4.3+4.5.3+...+2015.2016.3
=>1.2.(2+1)+2.3.(4-1)+3.4.(5-2)+...+2015.2016.(2017-2014)
=>1.2.2+1.2.1+2.3.4-2.3.1+3.4.5-3.4.2+...+2015.2016.2017-2015.2016.2014
=>2015.2016.2017