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\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)\)
\(A=1-\frac{1}{2^{2016}}< 1\)
a, \(2^4:2^3+3\left|10\right|-\left(-18\right)\)
\(=2+3\cdot10+18\)
\(=20+30=50\)
b, \(\frac{1}{2}-\frac{1}{3}+9\left(-\frac{1}{2}\right)^2-\frac{1}{-6}\)
\(=\frac{1}{2}-\frac{1}{3}+9\cdot\frac{1}{4}+\frac{1}{6}\)
\(=\frac{3}{6}-\frac{2}{6}+\frac{9}{4}+\frac{1}{6}\)
\(=\frac{1}{6}+\frac{1}{6}+\frac{9}{4}=\frac{1}{3}+\frac{9}{4}=\frac{4}{12}+\frac{27}{12}=\frac{31}{12}\)
B=2016/1+2015/2+...+1/2016
=1+(2015/2+1)+(2014/3+1)+...+(1/2016+1)
=2017/2+2017/3+...+2017/2016+2017/2017
=2017(1/2+1/3+...+1/2017)
=2017A
Vậy tỉ số giửa A và B là 1/2017
\(=\frac{12}{7}\cdot\frac{3}{4}-\frac{6}{7}\cdot\frac{4}{3}+\frac{6}{7}\)
\(=\frac{6}{7}\left(\frac{3}{2}-\frac{4}{3}+1\right)\)
\(=\frac{6}{7}\left(\frac{1}{6}+1\right)=\frac{6}{7}\cdot\frac{7}{6}=1\)
2.
\(=2017\cdot2018\cdot\left[\left(2016\cdot2018\right)-\left(2016\cdot2017\right)\right]\)
\(=2017\cdot2018\cdot2016\left(2018-2017\right)=2016\cdot2017\cdot2018\)
3.
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{100}-1\right)=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{99}{100}\)
\(=\frac{1}{100}\)
4.
\(=\frac{1+2+2^2+2^4+...+2^9}{2\left(1+2+2^2+2^3+2^4+...+2^9\right)}\)
\(=\frac{1}{2}\)
mình chỉ làm được câu 3 thôi
có \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)....\left(\frac{1}{100}-1\right)\)
\(=\frac{-1}{2}\times\frac{-2}{3}\times....\times\frac{-99}{100}\)
\(=\frac{\left(-1\right)\left(-2\right)....\left(-99\right)}{2\times3\times....\times100}\)
\(=\frac{-\left(1\times2\times....\times99\right)}{2\times3\times....\times100}\)
\(=\frac{-1}{100}\)
