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Ta có : \(A=1\frac{1}{2}\cdot1\frac{1}{3}\cdot1\frac{1}{4}\cdot...\cdot1\frac{1}{2015}\)
\(\Rightarrow A=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{2016}{2015}\Rightarrow A=\frac{3\cdot4\cdot5\cdot...\cdot2016}{2\cdot3\cdot4\cdot...\cdot2015}\)
\(\Rightarrow A=\frac{2016}{2}=1008\)
Vậy A = 1008
1) Ta có : \(\frac{x-2}{4}=\frac{5+x}{3}\)
\(\Rightarrow\left(x-2\right).3=\left(5+x\right).4\)
\(\Rightarrow3x-6=20+4x\)
\(\Rightarrow3x=26+4x\)
\(\Rightarrow3x=26+x+3x\)
\(\Rightarrow0=26+x\)
\(\Rightarrow x=0-26\)
\(\Rightarrow x=-26\)
2) Ta có : \(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\)
\(\Rightarrow\frac{1}{A}=1+2+2^2+...+2^{2012}\)
\(\Rightarrow\frac{2}{A}=2+2^2+2^3+...+2^{2013}\)
\(\Rightarrow\frac{2}{A}-\frac{1}{A}=\left(2+2^2+2^3+...+2^{2013}\right)-\left(1+2+2^2+...+2^{2012}\right)\)
\(\Rightarrow\frac{1}{A}=2^{2013}+1\)
\(\Rightarrow A=\frac{1}{2^{2013}+1}\)
\(D=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2014}{2015}\)
\(D=\frac{1\cdot2\cdot3\cdot...\cdot2014}{2\cdot3\cdot4\cdot...\cdot2015}=\frac{1}{2015}nhebn\)
(2/2-1/2).(3/3-1/3).(4/4-1/4)............(2015/2015-1/2015 )
1/2.2/3.3/4.....................2014/2015
=1/2015
a. \(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2014}}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{2013}}\)
\(\Rightarrow3A-A=1-\frac{1}{3^{2014}}\)
\(\Rightarrow2A=1-\frac{1}{3^{2014}}\)
\(\Rightarrow A=\left(1-\frac{1}{3^{2014}}\right):2=\frac{1}{2}-\frac{1}{3^{2014}.2}=\frac{3^{2014}-1}{3^{2014}.2}\)
b.\(B=\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2014}}\)
\(\Rightarrow2B=1+\frac{1}{2^2}+....+\frac{1}{2^{2013}}\)
\(\Rightarrow2B-B=1-\frac{1}{2^{2014}}\)
\(\Rightarrow B=1-\frac{1}{2^{2014}}\)
\(1\frac{1}{2}.1\frac{1}{3}.1\frac{1}{4}.....1\frac{1}{2015}\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}........\frac{2016}{2015}\)
\(=\frac{3.4.5.....2016}{2.3.4....2015}=\frac{2016}{2}=1008\)
\(A=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{2016}{2015}\)
\(A=\frac{2016}{2}=1008\)
Xong nhé bạn!