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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
Cho A = 1/2 + 1/3 + 1/4 + ... + 1/2017 B = 1/2015 + 2/2014 +3/2013 + ...+ 2015/2 + 2016/1 Tính B : A
Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
số con số trong phép tính là \(2016\) con số
\(A=-1-2+3+4-5-6+7+8-...-2013-2014+2015+2016\)
\(A=\left(-1-2+3+4\right)+\left(-5-6+7+8\right)+...+\left(-2013-2014+2015+2016\right)\)
\(A=4+4+4+...+4\)
ta có số nhóm trong phép tính là \(\dfrac{2016}{4}=504\)
\(\Rightarrow A=4.504=2016\) vậy \(A=2016\)
\(\dfrac{x-1}{2012}+\dfrac{x-2}{2013}+\dfrac{x-3}{2014}=\dfrac{x-4}{2015}+\dfrac{x-5}{2016}+\dfrac{x-6}{2017}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2012}+1\right)+\left(\dfrac{x-2}{2013}+1\right)+\left(\dfrac{x-3}{2014}+1\right)=\left(\dfrac{x-4}{2015}+1\right)+\left(\dfrac{x-5}{2016}+1\right)+\left(\dfrac{x-6}{2017}+1\right)\)
\(\Leftrightarrow\dfrac{x+2011}{2012}+\dfrac{x+2011}{2013}+\dfrac{x+2011}{2014}-\dfrac{x+2011}{2015}-\dfrac{x+2011}{2016}-\dfrac{x+2011}{2017}=0\)
\(\Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}-\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\)
\(\Leftrightarrow x=-2011\)( do \(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}-\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\ne0\))
Mẫu số = 2015/1 + 2014/2 + 2013/3 + ... + 1/2015
= (1 + 1 + ... + 1) + 2014/2 + 2013/3 + ... + 1/2015
2015 số 1
= (2014/2 + 1) + (2013/3 + 1) + ... + (1/2015 + 1) + 1
= 2016/2 + 2016/3 + ... + 2016/2015 + 2016/2016
= 2016 × (1/2 + 1/3 + ... + 1/2015 + 1/2016)
=> phân số đề bài cho = 1/2016
Ta có : \(\frac{x+5}{2012}+\frac{x+4}{2013}+\frac{x+3}{2014}=\frac{x+2}{2015}+\frac{x+1}{2016}+\frac{x}{2017}\)
\(\Rightarrow\frac{x+5}{2012}+1+\frac{x+4}{2013}+1+\frac{x+3}{2014}=\frac{x+2}{2015}+1+\frac{x+1}{2016}+1+\frac{x}{2017}+1\)
\(\Leftrightarrow\frac{x+2017}{2012}+\frac{x+2017}{2013}+\frac{x+2017}{2014}=\frac{x+2017}{2015}+\frac{x+2017}{2016}+\frac{x+2017}{2017}\)
\(\Leftrightarrow\frac{x+2017}{2012}+\frac{x+2017}{2013}+\frac{x+2017}{2014}-\frac{x+2017}{2015}-\frac{x+2017}{2016}-\frac{x+2017}{2017}=0\)
\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}-\frac{1}{2017}\right)=0\)
\(\text{Mà
}\)\(\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}-\frac{1}{2017}\right)\ne0\)
\(\text{Nên : }\) x + 2017 = 0
=> x = -2017
=>3A= 3^2017-3^2016+3^2015-...-3^2+3
=>3A+A=4A=3^2017+1=>A=\(\frac{3^{2017}+1}{4}\)
B tương tự nha