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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
a/ (x-2)/4 =5+x/2
=> \(\frac{x-2}{4}\) = \(\frac{x+5}{2}\)
=> 2(x-2) = 4(x+5)
2x-4 =4x+20
4x-2x=-20-4
2x=-24
x=-12
\(A=\frac{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=\frac{1+\left(1+\frac{2016}{2}\right)+\left(1+\frac{2015}{3}\right)+...+\left(1+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=\frac{\frac{2018}{2018}+\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2017}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=\frac{2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=2018\)
Ta có :
\(A=\frac{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=\frac{\left(\frac{2017}{1}-1-1-...-1\right)+\left(\frac{2016}{2}+1\right)+\left(\frac{2015}{3}+1\right)+...+\left(\frac{1}{2017}+1\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=\frac{\frac{2018}{2018}+\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=\frac{2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=2018\)
Vậy \(A=2018\)
Chúc bạn học tốt ~
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 : \(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\)
\(\Rightarrow\frac{1}{A}=1+2+2^2+...+2^{2012}\)
\(\Rightarrow\frac{2}{A}=2+2^2+2^3+...+2^{2013}\)
\(\Rightarrow\frac{2}{A}-\frac{1}{A}=\left(2+2^2+2^3+...+2^{2013}\right)-\left(1+2+2^2+...+2^{2012}\right)\)
\(\Rightarrow\frac{1}{A}=2^{2013}+1\)
\(\Rightarrow A=\frac{1}{2^{2013}+1}\)
1/1/2.1/1/3.1/1/4....1/1/2015=A
=)A=3/2.4/3.5/4....2016/1015
=)A=(3.4.5...2016)/(2.3.4...2015)
=)A=2016/2=1008
Vậy A =1008
\(D=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2014}{2015}\)
\(D=\frac{1\cdot2\cdot3\cdot...\cdot2014}{2\cdot3\cdot4\cdot...\cdot2015}=\frac{1}{2015}nhebn\)
(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, \(A=3a.2.b-a.432b-4ab\)
\(=6ab-432ab-4ab=-430ab\)
b, \(A=-430ab=\left(-430\right).\frac{1}{229}.\frac{1}{433}=\frac{-430}{229.433}\)
\(1\frac{1}{2}.1\frac{1}{3}.1\frac{1}{4}.....1\frac{1}{2015}\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}........\frac{2016}{2015}\)
\(=\frac{3.4.5.....2016}{2.3.4....2015}=\frac{2016}{2}=1008\)
\(A=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{2016}{2015}\)
\(A=\frac{2016}{2}=1008\)
Xong nhé bạn!