= 1000

= 10

a. \(2^2.2^3.2^{15}=2^{20}\)

\(3^{24}.3^{14}=3^{38}\)

\(c.5^{16}.5^9=5^{25}\)

\(d.3^{12}.3^{15}=3^{27}\)

a) 2⁴ . 2³ = 2⁷ = 128

b) 3³ . 3 . 3² = 3⁶ = 729

c) 100.10⁴ = 100 . 10000 = 1000000

d)5¹⁴ : 5¹² = 5² = 25

e) 7¹⁸ : 7¹⁶ = 7² = 49

f) 3^{9 :} 3⁵ = 3⁴ = 81

a) \(2^4.2^3=2^7=128\)

b) \(3^3.3.3^2=3^6=729\)

c) \(100.10^4=10^2.10^4=10^6=1000000\)

d) \(5^{14}:5^{12}=5^2=25\)

e) \(7^{18}:7^{16}=7^2=49\)

f) \(3^9:3^5=3^4=81\)

1+1+1+1+1+1+1+1+1+1=10

1+1+1+1+1+1+1+1+1-1=9

1+1+1+1+1+1+1+1-1-1=8

1+1+1+1+1+1+1-1-1-1=7

1+1+1+1+1+1-1-1-1-1=6

1+1+1+1+1-1-1-1-1-1=5

1+1+1+1-1-1-1-1-1-1=4

1+1+1-1-1-1-1-1-1-1=3

1+1-1-1-1-1-1-1-1-1=2

1-1-1-1-1-1-1-1-1-1-1=0

Nối cung ta điền dấu cộng hết

Các số còn lại ta giảm mỗi số một dấu cộng = dấu -

Úi khó thế em mới lớp 3

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)

\(< =>\frac{128}{256}+\frac{64}{256}+\frac{32}{256}+\frac{16}{256}+\frac{8}{256}+\frac{4}{256}+\frac{2}{256}+\frac{1}{256}\)

\(< =>\frac{128+64+32+16+8+4+2+1}{256}\)

\(< =>\frac{255}{256}\)

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

\(< =>\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(< =>\frac{1}{1}-\frac{1}{100}\)

\(< =>\frac{99}{100}\)

\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{100}\right)\)

\(< =>\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)

\(< =>\frac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\)

\(< =>\frac{1}{100}\)

mk chuc ban hoc tot nhe :))

1+1+11+11+1+1+1+1+1+1+1+1+1+111+1+1+1+1+1+1+1+1+1+1111+1+1+1+11+1+1+1+1+1+1+1+1+1+1=1288

= 1288 nhé !!!

\(\frac{1}{1.7}+\frac{1}{7.13}+\frac{1}{13.19}+...+\frac{1}{61.67}\)

=6.\(\left(\frac{1}{1.7}+\frac{1}{7.13}+...+\frac{1}{61.67}\right)\):6

=\((\frac{6}{1.7}+\frac{6}{7.13}+...+\frac{6}{61.67}):6\)

=\(\left(1-\frac{1}{7}+\frac{1}{7}+\frac{1}{13}+...+\frac{1}{61}+\frac{1}{67}\right):6\)

=\(\left(1-\frac{1}{67}\right):6\)

=\(\frac{66}{67}:6=\frac{66}{67}.\frac{1}{6}=\frac{11}{67}\)