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\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2016\cdot2017}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(=1-\frac{1}{2017}=\frac{2016}{2017}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2016.2017}\)
\(A=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+......+\left(\frac{1}{2016}-\frac{1}{2017}\right)\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2016}-\frac{1}{2017}\)
\(A=\frac{1}{1}-\frac{1}{2017}\)
\(A=\frac{2016}{2017}\)
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2016.2017}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow A=1-\frac{1}{2017}\)
\(\Rightarrow A=\frac{2016}{2017}\)
A=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2016}-\frac{1}{2017}\)
A=\(\frac{1}{1}-\frac{1}{2017}\)
A=\(\frac{2016}{2017}\)
\(\frac{33\cdot10^3}{2^3\cdot5\cdot10^3+7000}=\frac{33\cdot10^3}{40\cdot10^3+7\cdot10^3}=\frac{33\cdot10^3}{\left(40+7\right)\cdot10^3}=\frac{33}{47}\)
\(\frac{3774}{5217}=\frac{2\cdot3\cdot17\cdot37}{3\cdot37\cdot47}=\frac{2\cdot17}{47}=\frac{34}{47}\)
Vì 33 < 34 => \(\frac{33}{47}< \frac{34}{47}\)hay \(\frac{33\cdot10^3}{2^3\cdot5\cdot10^3+7000}< \frac{3774}{5217}\)
1/
+) \(\frac{3}{6}=\frac{2}{4};\frac{3}{2}=\frac{6}{4};\frac{4}{6}=\frac{2}{3};\frac{4}{2}=\frac{6}{3}\)
2/
\(A=\frac{3n-5}{n+4}=\frac{3n+12-17}{n+4}=\frac{3\left(n+4\right)}{n+4}-\frac{17}{n+4}=3-\frac{17}{n+4}\)
Để A nguyên <=> n + 4 thuộc Ư(17) = {1;-1;17;-17}
| n+4 | 1 | -1 | 17 | -17 |
| n | -3 | -5 | 13 | -21 |
Vậy...
3/
\(S=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(=1-\frac{1}{2017}\)
\(=\frac{2016}{2017}\)
\(A=\frac{3n+12-7}{n+4}=\frac{3\left(n+4\right)}{n+4}-\frac{7}{n+4}=3-\frac{7}{n+4}\)
=> n-4 \(\in\) Ư (7)
n-4=1
n=4+1=5
n-4=-1
n=-1+4=3
n-4=7
n=4+7=11
n-4=-7
n=-7+4=-3
a.\(\frac{2015.2016-1}{2015.2016}=1-\frac{1}{2015.2016}\)
\(\frac{2016.2017-1}{2016.2017}=1-\frac{1}{2016.2017}\)
vì \(\frac{1}{2015.2016}>\frac{1}{2016.2017}\)
=>\(-\frac{1}{2015.2016}< -\frac{1}{2016.2017}\)
=>\(1-\frac{1}{2015.2016}< 1-\frac{1}{2016.2017}\)