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\(5^x=125\)
\(5^x=5^3\)
=> x=3 ( vì cơ số 5>1)
\(3^2.x=81\)
\(9x=81\)
\(x=81:9\)
\(x=9\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1)
987 = 9.102 + 8.101 + 7.100
2564 = 2.103 + 5.102 + 6.101 + 4.100
abcde = a.104 + b.103 + c.102 + d.101 + e.100
2)
a) n = 1 b ) n = 0
3)
a) 13 + 23 = 1 + 8 = 9 = 32
b) 13 + 23 + 33 = 1 + 8 + 27 = 36 = 62
c ) 13 + 23 + 33 + 43 = 1 + 8 + 27 + 64 = 100 = 102
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a) Ta có
A = n / n+1 = 1-(1/n+1)
A = n+2 / n+3 = 1-(1/n+3)
Vì 1/n+1 > 1/n+3
=> n/n+1 < n+2/n+3
=> A<B
![](https://rs.olm.vn/images/avt/0.png?1311)
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Lời giải:
a.
\(\frac{n+1}{n+2}=\frac{n+1}{n+2}+1-1=\frac{2n+3}{n+2}-1\)
\(> \frac{2n+3}{n+3}-1=\frac{(n+3)+n}{n+3}-1=\frac{n}{n+3}\)
b.
\(10A=\frac{10^{12}-10}{10^{12}-1}=\frac{(10^{12}-1)-9}{10^{12}-1}=1-\frac{9}{10^{12}-1}<1\)
\(10B=\frac{10^{11}+10}{10^{11}+1}=\frac{(10^{11}+1)+9}{10^{11}+1}=1+\frac{9}{10^{11}+1}>1\)
$\Rightarrow 10A< 10B\Rightarrow A< B$
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