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a) ta có:
\(\frac{-1}{2}-1\le x\le\frac{1}{2}.3\)
hay \(-1,5\le x\le1,5\)
vì x\(\in Z\) nên ta chọn x=-1,0,1
ta có:
3S=\(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\)
3S-S=\(\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^9}\right)\)
2S=1-\(\frac{1}{3^9}\)
s=\(\left(1-\frac{1}{3^9}\right):2\)
a. Nhân 2 vế của S với 3 rồi cộng S và 3S. Rút gọn sẽ ra kết quả
a) \(\frac{7}{11}-\left(\frac{3}{5}+\frac{7}{11}\right)=-\frac{3}{5}\)
b) \(\left(\frac{11}{22}+\frac{5}{11}\right)-\frac{19}{22}=\frac{1}{11}\)
c) \(\frac{2}{9}.\frac{4}{5}+\frac{2}{9}.\frac{14}{5}=\frac{4}{5}\)
d) \(-\frac{3}{2}.\frac{7}{10}-\frac{3}{2}.\frac{1}{10}=-\frac{6}{5}\)
e) \(\left(0,75-1+\frac{1}{4}\right):\left(\frac{1515}{1616}+\frac{1616}{1717}\right)=0\)
Ta có :
\(S=2015+\frac{2015}{1+2}+\frac{2015}{1+2+3}+...+\frac{2015}{1+2+3+..+2016}\)
\(=2015.\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+..+2016}\right)\)
\(=2015.\left(1+\frac{1}{\frac{\left(2+1\right).2}{2}}+\frac{1}{\frac{\left(3+1\right).3}{2}}+...+\frac{1}{\frac{\left(2016+1\right).2016}{2}}\right)\)
\(=2015.\left(\frac{2}{2}+\frac{2}{2.\left(2+1\right)}+\frac{2}{3.\left(3+1\right)}+...+\frac{2}{2016.\left(2016+1\right)}\right)\)
\(=2015.2.\left(\frac{1}{2}+\frac{1}{2.\left(2+1\right)}+\frac{1}{3.\left(3+1\right)}+...+\frac{1}{2016.\left(2016+1\right)}\right)\)
\(=2015.2.\left(\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\right)\)
\(=2015.2.\left(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)
\(=2015.2.\left(\frac{1}{2}+\frac{1}{2}-\frac{1}{2017}\right)\)
\(=2015.2.\left(1-\frac{1}{2017}\right)\)
\(=2015.2.\frac{2016}{2017}\)
=\(\frac{2015.2.2016}{2017}\)
=\(\frac{8124480}{2017}\)
Vậy \(S=\frac{8124480}{2017}\)
Bài 2 :
Ta có :
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{16}\right)+\left(\frac{1}{16}-\frac{1}{32}\right)\)
\(=1-\frac{1}{32}\)
và \(B=\frac{2003}{2004}=1-\frac{1}{2004}\)
Vì \(\frac{1}{32}>\frac{1}{2004}\) nên A < B