D

D. \(2^3.2^2=2^6\)

D

Ta có: \(S=20.21+21.22+22.23+....+39.40\)

\(\Rightarrow3S=3.20.21+3.21.22+.....+3.39.40\)

\(\Rightarrow3S=20.21.\left(22-19\right)+22.23.\left(24-21\right)+.....+39.40.\left(41-38\right)\)

\(\Rightarrow3S=39.40.41-19.20.21\)

\(\Rightarrow3S=55980\)

\(\Rightarrow S=18660\)

Cách giải

=(39.40-20.21):0.01+1

=19.19:0.01+1

=1919+1

=1920

3s=20.21.(22-19)+21.22.(23-20)+22.23.(24-21)+...+39.40.(41-38)

=20.21.22 - 20.21.19+21.22.23 - 21.22.20 +22.23.24 -22.23.21 + ...+ 39.40.41-39.40.38

=39.40.41- 20.21.19=63960-7980=55980

s=18660

\(S=20.21+22.23+23.24+....+39.40\)
\(3S=3.20.21+3.22.23+3.23.24+...+3.39.40\)
\(3S=20.21.\left(22-19\right)+22.23.\left(24-21\right)+....+39.40.\left(41-38\right)\)
\(3S=39.40.41-19.20.21\)
\(3S=55980\)
\(S=18660\).

a,  15 . { 32 : [ 6 - 5 + 5 ( 9 : 3 ) ] + 3 } - 2018 0

= 15.{32:[1+15]+3}–1

= 15.5–1

= 74

b,  25 . { 2 7 : [ 12 - 4 + 2 2 . 16 : 2 3 ] - 2 4 }

= 25.{128:[8+4.2]–16}

= 25.24

= 600

c,  2019 . { 101 - 1000 : [ 2 2 . 2 3 + 5 6 : 5 3 - 6 2 : 11 - 2018 0 ] }

= 2019.{101–1000:[(32+125–36):11–1]}

= 2019.{101–1000:[121:11–1]}

= 2019.{101–1000:10}

= 2019.1

= 2019

đặt S=20.21+22.23+23.24+....+38.39+39.40

3s = 3.20.21+3.22.23+3.23.24+...+3.38.39+3.39.40

=20.21.(22-19)+22.23.(24-21)+23.24.(25-22)+...+39.40.(41-38)

=39.40.41-19.20.21=55980

suy ra s = 18660

Bài 1 :

\(A=2+2^2+2^3+2^4+...+2^{118}+2^{119}+2^{120}\)

\(\Rightarrow A=2\left(1+2^{ }+2^2\right)+2^4\left(1+2^{ }+2^2\right)+...+2^{118}\left(1+2^{ }+2^2\right)\)

\(\Rightarrow A=2.7+2^4.7+...+2^{118}.7\)

\(\Rightarrow A=7.\left(2+2^4+...+2^{118}\right)⋮7\)

\(\Rightarrow dpcm\)

Bài 2 :

\(...=23\left(78+22\right)-15=23.100-15=2300-15=2285\)

ko bt bài 1

bài 2 là

=23.(78+22)-15

=23.100-15

=2300-15

=2275

hết

Bài 1 :

A = (2 + 2² + 2³) + (2⁴ + 2⁵ + 2⁶) + ... + (2¹¹⁸ + 2¹¹⁹ + 2¹²⁰)

A = 2.(1 + 2 + 4) + 2⁴.(1 + 2 + 4) + ... + 2¹¹⁸.(1 + 2 + 4)

A = 7.(2 + 2⁴ + ...  + 2¹¹⁸)   ⋮  7 (đfcm)