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\(A=\dfrac{7}{10}+\dfrac{7}{10^2}+\dfrac{7}{10^3}+...+\dfrac{7}{10^{2011}}\)
\(\Rightarrow10A=7+\dfrac{7}{10}+\dfrac{7}{10^2}+...+\dfrac{7}{10^{2010}}\)
\(\Rightarrow10A-A=7+\dfrac{7}{10}+\dfrac{7}{10^2}+...+\dfrac{7}{10^{2010}}-\left(\dfrac{7}{10}+\dfrac{7}{10^2}+\dfrac{7}{10^3}+...+\dfrac{7}{10^{2011}}\right)\)
\(\Rightarrow9A=7-\dfrac{7}{10^{2011}}\)
\(\Rightarrow A=\dfrac{7}{9}.\left(1-\dfrac{1}{10^{2011}}\right)\)
\(10A=7+\frac{7}{10}+\frac{7}{10^2}+...+\frac{7}{10^{99}}\)
\(10A-A=\left(7+\frac{7}{10}+...+\frac{7}{10^{99}}\right)-\left(\frac{7}{10}+\frac{7}{10^2}+...+\frac{7}{10^{100}}\right)\)
\(9A=7-\frac{7}{10^{100}}\)
\(A=\frac{7-\frac{7}{10^{100}}}{9}\)
Lời giải chi tiết:
2 = 1 + 1 |
6 = 2 + 4 |
8 = 5 + 3 |
10 = 8 + 2 |
3 = 1 + 2 |
6 = 3 + 3 |
8 = 4 + 4 |
10 = 7 + 3 |
4 = 3 + 1 |
7 = 6 + 1 |
9 = 8 + 1 |
10 = 6 + 4 |
4 = 2 + 2 |
7 = 5 + 2 |
9 = 7 + 2 |
10 = 5 + 5 |
5 = 4 + 1 |
7 = 4 + 3 |
9 = 6 + 3 |
10 = 10 + 0 |
5 = 3 + 2 |
8 = 7 + 1 |
9 = 5+ 4 |
10 = 0 + 10 |
6 = 5 + 1 |
8 = 6 + 2 |
10 = 9 + 1 |
1 = 0 + 1 |
2=1+1 6=2+4 8=5+3 10=8+2
3=1+2 6=3+3 8=4+4 10=7+3
4=3+1 7=6+1 9=8+1 10=6+4
4=2+2 7=5+2 9=7=2 10=5+5
5=4+1 7=4+3 9=6+3 10=10+0
5=3+2 8=7+1 9=5=4 10=0+10
6=5+1 8=6=2 10=9+1 1=0+1
Lời giải chi tiết
| 9 + 1 = 10 | 8 + 2 = 10 | 7 + 3 = 10 | 6 + 4 = 10 |
| 5 + 5 = 10 | 1 + 9 = 10 | 2 + 8 = 10 | 3 + 7 = 10 |
| 4 + 6 = 10 | 10 – 5 = 5 | 10 – 1 = 9 | 10 – 2 = 8 |
| 10 – 3 = 7 | 10 – 4 = 6 | 10 – 0 = 10 | 10 – 9 = 1 |
| 10 – 8 = 2 | 10 – 7 = 3 | 10 – 6 = 4 | 10 – 10 = 0 |
1 + 9 = 10 2 + 8 = 10 3 + 7 = 10 4 + 6 = 10 5 + 5 = 10
10 - 1 = 9 10 - 2 = 8 10 - 3 = 7 10 - 4 = 6 10 - 5 = 5
6 + 4 = 10 7 + 3 = 10 8 + 2 = 10 9 + 1 = 10 10 + 0 = 10
10 - 6 = 4 10 - 7 = 3 10 - 8 = 2 10 - 9 = 1 10 - 0 = 10
Ta có \(A=\frac{10}{2^7}+\frac{10}{2^6}=\frac{5}{2^6}+\frac{10}{2^6}=\frac{15}{2^6}\)
Lại có B = \(\frac{11}{2^7}+\frac{9}{2^6}=\frac{5,5}{2^6}+\frac{9}{2^6}=\frac{14,5}{2^6}\)
Vì \(\frac{15}{2^6}>\frac{14,5}{2^6}\Rightarrow A>B\)
b) Ta có : \(A=\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}=\frac{-70}{10^{2006}}+\frac{-15}{10^{2006}}=\frac{-85}{10^{2006}}\)
Lại có B = \(\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}=\frac{-150}{10^{2006}}+\frac{-7}{10^{2006}}=\frac{-157}{10^{2006}}\)
Vì \(\frac{-85}{10^{2006}}>\frac{-157}{10^{2006}}\Rightarrow A< B\)
Chọn câu C: \(\frac{5}{2};\frac{7}{4};\frac{4}{5};\frac{7}{10}\)
\(\left(\dfrac{2}{7}\right)^{10}.7^{10}=\dfrac{2^{10}}{7^{10}}.7^{10}=2^{10}=1024\)