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Bài 1:
a, \(64^2=\left(2^6\right)^2=2^{12}\)
\(32^5=\left(2^5\right)^5=2^{25}\)
\(2^{12}< 2^{25}\Rightarrow64^2< 32^5\)
b, \(10^5=\left(5.2\right)^5=5^5.2^5\)
\(5^{10}=5^{5+5}=5^5.5^5\)
\(5^5.2^5< 5^5.5^5\Rightarrow10^5< 5^{10}\)
c, 256^ mấy thế hả bạn?
d, \(9^{200}=\left(9^2\right)^{100}=81^{100}\)
\(81^{100}>10^{100}\Rightarrow9^{200}>10^{100}\)
Bài 2:
\(9.27^2.81^3.216\)
\(=3^2.\left(3^3\right)^2.\left(3^4\right)^3.2^3.3^3\)
\(=3^2.3^6.3^{12}.3^3.2^3\)
\(=3^{2+6+12+3}.2^3\)
\(=3^{23}.2^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(199^{20}=\left(199^4\right)^5=1568239201^5\)
\(2003^{15}=\left(2003^3\right)^5=8036054027^5\)
Vì \(8036054027^5>1568239201^5\) nên \(2003^{15}>199^{20}\)
Ta có :
19920 < 21620 = ( 63 )20 = 660
200315 > 129615 = ( 64 )15 = 660
Vì 19920 < 660 < 200315
=> 19920 < 200315
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1: 8=2^3
2: 25=5^2
3: 4=2^2
4: 49=7^2
5: 81=9^2
6: 36=6^2
7: 100=10^2
8: 121=11^2
9: 144=12^2
10: 169=13^2
11: 27=3^3
12: 125=5^3
13: 1000=10^3
14: 32=2^5
15: 243=3^5
16: 343=7^3
17: 216=6^3
18: 64=4^3
19: 225=15^2
20: 128=2^7
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1111=1000+100+10+1=10^3+10^2+10^1+1\)
\(1247=1000+2\times100+4\times10+7=10^3+2\times10^2+4\times10^1+7\)
\(1543=1000+5\times100+4\times10+3=10^3+5\times10^2+4\times10^1+3\)
\(1794=1000+7\times100+9\times10+4=10^3+7\times10^2+9\times10^1+4\)
\(1876=1000+8\times100+7\times10+6=10^3+8\times10^2+7\times10^1+6\)
\(1008=1000+8=10^3+8\)
\(1080=1000+8\times10=10^3+8\times10^1\)
\(1709=1000+7\times100+9=10^3+7\times10^2+9\)
\(1500=1000+5\times100=10^3+5\times10^2\)
\(1097=1000+9\times10+7=10^3+9\times10^1+7\)
\(20000=2\times10000=2\times10^4\)
\(27000=2\times10000+7\times1000=2\times10^4+7\times10^3\)
\(28700=2\times10000+8\times1000+7\times100\)
\(24004=2\times10000+4\times1000+4=2\times10^4+4\times10^3+4\)
\(26013=2\times10000+6\times1000+1\times10+3=2\times10^4+6\times10^3+10^1+3\)
\(20508=2\times10000+5\times100+8=2\times10^4+5\times10^2+8\)
\(20358=2\times10000+3\times100+5\times10+8=2\times10^4+3\times10^2+5\times10^1+8\)
\(20009=2\times10000+9=2\times10^4+9\)
\(29830=2\times10000+9\times1000+8\times100+3\times10=2\times10^4+9\times10^3+8\times10^2+3\times10^1\)
\(25763=2\times10000+5\times1000+7\times100+6\times10+3=2\times10^4+5\times10^3+7\times10^2+6\times10^1+3\)
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Bài 1:
19920 < 20020 = (23 . 52)20 = 260 . 540
200315 > 200015 = (24 . 53)15 = 260 . 545
Do: 260 . 540 < 260 . 545 => 19920 < 260 . 540 < 260 . 545 < 200315 => 19920 < 200315
Bài 2:
a/ (3 . x - 9) . 312 = 315 => 3 . x - 9 = 315 : 312
=> 3 . x - 9 = 27 => 3 . x = 27 + 9
=> 3 . x = 36 => x = 12
b/ (7 . x + 6) . 55 = 58 => 7 . x + 6 = 58 : 55
=> 7 . x + 6 = 125 => 7 . x = 125 - 6
=> 7 . x = 119 => x = 17
c/ (x - 5)4 = (x - 5)6
<=> x - 5 = 1 hoặc x - 5 = -1 hoặc x - 5 = 0
=> x = 6 hoặc x = 4 hoặc x = 5
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a) \(4\frac{7}{10}< 6\frac{7}{10}\)(4 < 6)
b) \(3\frac{4}{15}< 3\frac{11}{15}\)(4/15 < 11/15)
c) \(5\frac{1}{9}>2\frac{2}{5}\)(5 > 2)
d) \(2\frac{2}{3}=2\frac{10}{15}\)(10/15 = 2/3)
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Bài 1:
16 = 24
64 = 26
1 = 20
1/32 = 2-5
1/8 = 2-3
0,5 = 1/2 = 2-1
0,25 = 1/4 = 2-2
Bài 2:
a) ta có: 1624 = (42)24 = 448
2716 = (33)16 = 348 < 448
=> 1624>2716
b) ta có: 1020 = (102)10 = 10010 > 9010
1715 và 259
ta có:
1715>1615 ; 1615= (24)15=260
Vì 260> 259=>1615>259
=>1715>259