1. Số hạt thóc có trong ô thứ 7 của bàn cờ nói trong bài toán mở đầu:

2.2.2.2.2.2 = 2⁶ = 64

2. a) 23 197 = 2.10⁴  + 3.10³  + 1.10²  + 9.10¹  + 7

    b) 203 184 = 2.10⁵  + 0.10⁴  + 3.10³ + 1.10² + 8.10¹ + 4

\(215=2\cdot10^2+1\cdot10+5\cdot10^0\)

\(902=9\cdot10^2+2\cdot10^0\)

\(2020=2\cdot10^3+2\cdot10\)

\(883011=8\cdot10^5+8\cdot10^4+3\cdot10^3+10+1\cdot10^0\)

\(215 = 2. 10^2 + 1. 10^1 + 5. 10^0\)

\(902 = 9.10^2 + 0.10^1+ 2 . 10^0\)

\(2 020 = 2.10^3+0.10^2+2.10^1+0. 10^0\)

\(883 001 = 8.10^5+8.10^4+3.10^3+0.10^2+0.10^1+1. 10^0\)

215=2.10² + 1.10 + 5

 902 = 9.10² + 0.10 + 2

2 020 = 2.10³ + 2.10 + 0

883 001 = 8.10⁵ + 8.10⁴ + 3.10³ + 1

Chúc bạn học tốt

$a)$

$17120=10000+7.1000+100+2.10$

$={10}^4+7.{10}^3+{10}^2+2.10$

$b)$

$917111=9.100000+10000+7.1000+100+10+1$

$=9.{10}^5+{10}^4+7.{10}^3+{10}^2+10+1$

$c)$ với a khác 0

$\overline{abc}=a.100+b.10+c$

$=a{.10}^2+b.10+c$

$a)$

$17120=10000+7.1000+100+2.10$

$={10}^4+7.{10}^3+{10}^2+2.10$

$b)$

$917111=9.100000+10000+7.1000+100+10+1$

$=9.{10}^5+{10}^4+7.{10}^3+{10}^2+10+1$

$c)$ với a khác 0

$\overline{abc}=a.100+b.10+c$

$=a{.10}^2+b.10+c$

a, 17120 = 1.10⁴ + 7.10³ + 1.10² + 2.10¹ + 0

b, 917111 = 9.10⁵ + 1.10⁴ + 7.10³ + 1.10² + 1.10¹ + 1

c, \(\overline{abc}\) = \(a.10^2\) + \(b\).10 + c

\(\hept{215=2\times10^2+10+5\times10^0}\)

\(902=9\times10^2+2\times10^0\)

\(2020=2\times10^3+2\times10\)

\(883001=8\times10^5+8\times10^4+3\times10^3+10^0\)

215=2.10² + 1.10 + 5

 902 = 9.10² + 0.10 + 2

2 020 = 2.10³ + 2.10 + 0

883 001 = 8.10⁵ + 8.10⁴ + 3.10³ + 1

Chúc bạn học tốt

mnki mmmmmmmmmmmmmmmmmmmmm                               kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk

lam theo lua thua cac so la ra

a) 215=2.10^2+1.10+5

b) 902=9.10^2+2

c) 2020=2.10^3+2.10

d) 883001=8.10^5+8.10^4+3.10^3+1

\(a)\)\(215=2.10^2+1.10+5\)

\(b)\)\(902=9.10^2+2\)

\(c)\)\(2020=2.10^3+2.10\)

\(d)\)\(883001=8.10^5+8.10^4+3.10^3+1\)

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