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1.
\(\left(x+2\right)^3=\frac{1}{8}\)
\(\Rightarrow\left(x+2\right)^3=\left(\frac{1}{2}\right)^3\)
\(\Rightarrow x+2=\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{2}-2\)
\(\Rightarrow x=-\frac{3}{2}\)
Vậy \(x=-\frac{3}{2}.\)
2.
b) Ta có:
\(5^5-5^4+5^3\)
\(=5^3.\left(5^2-5+1\right)\)
\(=5^3.\left(25-5+1\right)\)
\(=5^3.21\)
Vì \(21⋮7\) nên \(5^3.21⋮7.\)
\(\Rightarrow5^5-5^4+5^3⋮7\left(đpcm\right).\)
c) Ta có:
\(2^{19}+2^{21}+2^{22}\)
\(=2^{19}.\left(1+2^2+2^3\right)\)
\(=2^{19}.\left(1+4+8\right)\)
\(=2^{19}.13\)
Vì \(13⋮13\) nên \(2^{19}.13⋮13.\)
\(\Rightarrow2^{19}+2^{21}+2^{22}⋮13\left(đpcm\right).\)
Chúc bạn học tốt!
A = 2100- 299 + 298 - 297 + ... + 22 - 2
=> 2A = 2101 - 2100 + 299 - 298 + ... + 23 - 22
Khi đó 2A + A = (2101 - 2100 + 299 - 298 + ... + 23 - 22) + (2100- 299 + 298 - 297 + ... + 22 - 2)
=> 3A = 2101 - 2
=> \(A=\frac{2^{201}-2}{3}\)
b) Ta có B = 3100- 399 + 398 - 397 + ... + 32 - 3 + 1
=> 3B = 3101 - 3100 + 399 - 398 + ... + 33 - 32 + 3
Khi đó 3B + B = (3101 - 3100 + 399 - 398 + ... + 33 - 32 + 3) + (3100- 399 + 398 - 397 + ... + 32 - 3 + 1)
=> 4B = 3101 + 1
=> B = \(\frac{3^{101}+1}{4}\)
a) \(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
=> \(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
=> \(2A+A=\left(2^{101}-2^{100}+...-2^2\right)+\left(2^{100}-2^{99}+...-2\right)\)
<=> \(3A=2^{101}-2\)
=> \(A=\frac{2^{101}-2}{3}\)
b) \(B=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\)
=> \(3A=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2+3\)
=> \(3A+A=\left(3^{101}-3^{100}+...+3\right)+\left(3^{100}-3^{99}+...+1\right)\)
<=> \(4A=3^{101}+1\)
=> \(A=\frac{3^{101}+1}{4}\)
Với \(x=100\)\(\Rightarrow x-1=99\)
Ta có: \(C=99+99x+99x^2+99x^3+.......+99x^n+99x^{n+1}\)
\(=x-1+\left(x-1\right).x+\left(x-1\right).x^2+........+\left(x-1\right).x^n+\left(x-1\right).x^{n+1}\)
\(=x-1+x^2-x+x^3-x^2+......+x^{n+1}-x^n+x^{n+2}-x^{n+1}\)
\(=-1+x^{n+2}=x^{n+2}-1\)
Thay \(x=100\)vào biểu thức ta được:
\(C=100^{n+2}-1\)
\(B=2^3+4^3+6^3+....+200^3\)
\(=\left(1.2\right)^3+\left(2.2\right)^3+\left(2.3\right)^3+...+\left(2.100\right)^3\)
\(=1^3.2^3+2^3.2^3+2^3.3^3+....+2^3.100^3\)
\(=1^3.8+2^3.8+3^3.8+....+100^3.8\)
\(=8\left(1^3+2^3+3^3+....+100^3\right)\)
\(\Rightarrow\frac{B}{A}=\frac{8\left(1^3+2^3+3^3+....+100^3\right)}{1^3+2^3+3^3+....+100^3}=8\)
Câu 1: Dân số thế giới tăng nhanh trong khoảng thời gian nào?
a. Trước Công nguyên b. Từ Công Nguyên- thế kỉ XI
c. Từ thế kỉ XIX- thế kỉ XX d. Từ thế kỉ XIX- nay
Chọn C
Câu 2: Những năm 50 của thế kỉ XX bùng nổ dân số diễn ra ở
a. Châu Âu, Á, Đại dương b. Châu Á,Phi và Mĩ La Tinh
c. Châu Mĩ, Đại dương, Phi. d. Châu Mĩ La Tinh, Á, Âu
Chọn B
b)
B=1x2+2x3+3x4+...+99x100
1/B=1/(1x2)+1/(2x3)+1/(3x4)+...+1/(99x100)
1/B=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/99-1/100
1/B=1/1-1/100
1/B=99/100
vì 1/B=99/100=>99.B=100
B=100/99
Vậy B=100/99
a) Ta có : A = 3 + 32 + 33 + 34 + ... + 399 + 3100
= (3 + 32) + (33 + 34) + ... + (399 + 3100)
= (3 + 32) + 32.(3 + 32) + .... + 398.(3 + 32)
= 12 + 32.12 + .... + 398.12
= 12.(1 + 32 + ... + 398) (1)
= 3.4.(1 + 32 + ... + 398) \(⋮\) 4
=> \(A⋮4\)
Từ (1) \(\Rightarrow A⋮12\)
b) B = 1 x 2 + 2 x 3 + 3 x 4 + ... + 99 x 100
3B = 1 x 2 x 3 + 2 x 3 x 3 + 3 x 4 x 3 + ... + 99 x 100 x 3
= 1 x 2 x 3 + 2 x 3 x (4 - 1) + 3 x 4 x (5 - 2) + .... + 99 x 100 x (101 - 98)
= 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5 - 2 x 3 x 4 + .... + 99 x 100 x 101 - 98 x 99 x 100
= 99 x 100 x 101 = 999 900
=> B = 333 300
c) Ta có : C = 12 + 22 + 32 + ... + 992 + 1002
= 1.1 + 2.2 + 3.3 + ... + 99.99 + 100.100
= 1.(2 - 1) + 2.(3 - 1) + 3.(4 - 1) + .... + 99.(100 - 1) + 100.(101 - 1)
= 1.2 - 1 + 2.3 - 2 + 3.4 - 3 + ... + 99.100 - 99 + 100.101 - 100
= (1.2 + 2.3 + 3.4 + ... + 99.100 + 100.101) - (1 + 2 + 3 + 4 + ... + 99 + 100)
Đặt B = 1.2 + 2.3 + 3.4 + ... + 99.100 + 100.101
=> 3B = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3 + 100.101.3
= 1.2.3 + 2.3(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98) + 100.101.(102 - 99)
= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100 + 100.101.102 - 99.100.101
= 100.101.102
= 1 030 200
=> B = 343 400
Khi đó : C = B - (1 + 2 + 3 + 4 + ... + 99 + 100)
= 343 400 - [(100 - 1) : 1 + 1] . (100 + 1) : 2
= 343 400 - 100 . 101 : 2
= 343 400 + 5050
= 348 450
Vậy C = 348 500
a) Ta có : A = 3 + 32 + 33 + 34 + ... + 399 + 3100
= (3 + 32) + (33 + 34) + ... + (399 + 3100)
= (3 + 32) + 32.(3 + 32) + .... + 398.(3 + 32)
= 12 + 32.12 + .... + 398.12
= 12.(1 + 32 + ... + 398) (1)
= 3.4.(1 + 32 + ... + 398) \(⋮\) 4
=> \(A⋮4\)
Từ (1) \(\Rightarrow A⋮12\)
b) B = 1 x 2 + 2 x 3 + 3 x 4 + ... + 99 x 100
3B = 1 x 2 x 3 + 2 x 3 x 3 + 3 x 4 x 3 + ... + 99 x 100 x 3
= 1 x 2 x 3 + 2 x 3 x (4 - 1) + 3 x 4 x (5 - 2) + .... + 99 x 100 x (101 - 98)
= 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5 - 2 x 3 x 4 + .... + 99 x 100 x 101 - 98 x 99 x 100
= 99 x 100 x 101 = 999 900
=> B = 333 300
c) Ta có : C = 12 + 22 + 32 + ... + 992 + 1002
= 1.1 + 2.2 + 3.3 + ... + 99.99 + 100.100
= 1.(2 - 1) + 2.(3 - 1) + 3.(4 - 1) + .... + 99.(100 - 1) + 100.(101 - 1)
= 1.2 - 1 + 2.3 - 2 + 3.4 - 3 + ... + 99.100 - 99 + 100.101 - 100
= (1.2 + 2.3 + 3.4 + ... + 99.100 + 100.101) - (1 + 2 + 3 + 4 + ... + 99 + 100)
Đặt B = 1.2 + 2.3 + 3.4 + ... + 99.100 + 100.101
=> 3B = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3 + 100.101.3
= 1.2.3 + 2.3(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98) + 100.101.(102 - 99)
= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100 + 100.101.102 - 99.100.101
= 100.101.102
= 1 030 200
=> B = 343 400
Khi đó : C = B - (1 + 2 + 3 + 4 + ... + 99 + 100)
= 343 400 - [(100 - 1) : 1 + 1] . (100 + 1) : 2
= 343 400 - 100 . 101 : 2
= 343 400 + 5050
= 348 450
Vậy C = 348 500
a) Ta có : A = 3 + 32 + 33 + 34 + ... + 399 + 3100
= (3 + 32) + (33 + 34) + ... + (399 + 3100)
= (3 + 32) + 32.(3 + 32) + .... + 398.(3 + 32)
= 12 + 32.12 + .... + 398.12
= 12.(1 + 32 + ... + 398) (1)
= 3.4.(1 + 32 + ... + 398) \(⋮\) 4
=> \(A⋮4\)
Từ (1) \(\Rightarrow A⋮12\)
b) B = 1 x 2 + 2 x 3 + 3 x 4 + ... + 99 x 100
3B = 1 x 2 x 3 + 2 x 3 x 3 + 3 x 4 x 3 + ... + 99 x 100 x 3
= 1 x 2 x 3 + 2 x 3 x (4 - 1) + 3 x 4 x (5 - 2) + .... + 99 x 100 x (101 - 98)
= 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5 - 2 x 3 x 4 + .... + 99 x 100 x 101 - 98 x 99 x 100
= 99 x 100 x 101 = 999 900
=> B = 333 300
c) Ta có : C = 12 + 22 + 32 + ... + 992 + 1002
= 1.1 + 2.2 + 3.3 + ... + 99.99 + 100.100
= 1.(2 - 1) + 2.(3 - 1) + 3.(4 - 1) + .... + 99.(100 - 1) + 100.(101 - 1)
= 1.2 - 1 + 2.3 - 2 + 3.4 - 3 + ... + 99.100 - 99 + 100.101 - 100
= (1.2 + 2.3 + 3.4 + ... + 99.100 + 100.101) - (1 + 2 + 3 + 4 + ... + 99 + 100)
Đặt B = 1.2 + 2.3 + 3.4 + ... + 99.100 + 100.101
=> 3B = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3 + 100.101.3
= 1.2.3 + 2.3(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98) + 100.101.(102 - 99)
= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100 + 100.101.102 - 99.100.101
= 100.101.102
= 1 030 200
=> B = 343 400
Khi đó : C = B - (1 + 2 + 3 + 4 + ... + 99 + 100)
= 343 400 - [(100 - 1) : 1 + 1] . (100 + 1) : 2
= 343 400 - 100 . 101 : 2
= 343 400 + 5050
= 348 450
Vậy C = 348 500
a) Ta có : A = 3 + 32 + 33 + 34 + ... + 399 + 3100
= (3 + 32) + (33 + 34) + ... + (399 + 3100)
= (3 + 32) + 32.(3 + 32) + .... + 398.(3 + 32)
= 12 + 32.12 + .... + 398.12
= 12.(1 + 32 + ... + 398) (1)
= 3.4.(1 + 32 + ... + 398) \(⋮\) 4
=> \(A⋮4\)
Từ (1) \(\Rightarrow A⋮12\)
b) B = 1 x 2 + 2 x 3 + 3 x 4 + ... + 99 x 100
3B = 1 x 2 x 3 + 2 x 3 x 3 + 3 x 4 x 3 + ... + 99 x 100 x 3
= 1 x 2 x 3 + 2 x 3 x (4 - 1) + 3 x 4 x (5 - 2) + .... + 99 x 100 x (101 - 98)
= 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5 - 2 x 3 x 4 + .... + 99 x 100 x 101 - 98 x 99 x 100
= 99 x 100 x 101 = 999 900
=> B = 333 300
c) Ta có : C = 12 + 22 + 32 + ... + 992 + 1002
= 1.1 + 2.2 + 3.3 + ... + 99.99 + 100.100
= 1.(2 - 1) + 2.(3 - 1) + 3.(4 - 1) + .... + 99.(100 - 1) + 100.(101 - 1)
= 1.2 - 1 + 2.3 - 2 + 3.4 - 3 + ... + 99.100 - 99 + 100.101 - 100
= (1.2 + 2.3 + 3.4 + ... + 99.100 + 100.101) - (1 + 2 + 3 + 4 + ... + 99 + 100)
Đặt B = 1.2 + 2.3 + 3.4 + ... + 99.100 + 100.101
=> 3B = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3 + 100.101.3
= 1.2.3 + 2.3(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98) + 100.101.(102 - 99)
= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100 + 100.101.102 - 99.100.101
= 100.101.102
= 1 030 200
=> B = 343 400
Khi đó : C = B - (1 + 2 + 3 + 4 + ... + 99 + 100)
= 343 400 - [(100 - 1) : 1 + 1] . (100 + 1) : 2
= 343 400 - 100 . 101 : 2
= 343 400 + 5050
= 348 450
Vậy C = 348 500
a) Ta có : A = 3 + 32 + 33 + 34 + ... + 399 + 3100
= (3 + 32) + (33 + 34) + ... + (399 + 3100)
= (3 + 32) + 32.(3 + 32) + .... + 398.(3 + 32)
= 12 + 32.12 + .... + 398.12
= 12.(1 + 32 + ... + 398) (1)
= 3.4.(1 + 32 + ... + 398) \(⋮\) 4
=> \(A⋮4\)
Từ (1) \(\Rightarrow A⋮12\)
b) B = 1 x 2 + 2 x 3 + 3 x 4 + ... + 99 x 100
3B = 1 x 2 x 3 + 2 x 3 x 3 + 3 x 4 x 3 + ... + 99 x 100 x 3
= 1 x 2 x 3 + 2 x 3 x (4 - 1) + 3 x 4 x (5 - 2) + .... + 99 x 100 x (101 - 98)
= 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5 - 2 x 3 x 4 + .... + 99 x 100 x 101 - 98 x 99 x 100
= 99 x 100 x 101 = 999 900
=> B = 333 300
c) Ta có : C = 12 + 22 + 32 + ... + 992 + 1002
= 1.1 + 2.2 + 3.3 + ... + 99.99 + 100.100
= 1.(2 - 1) + 2.(3 - 1) + 3.(4 - 1) + .... + 99.(100 - 1) + 100.(101 - 1)
= 1.2 - 1 + 2.3 - 2 + 3.4 - 3 + ... + 99.100 - 99 + 100.101 - 100
= (1.2 + 2.3 + 3.4 + ... + 99.100 + 100.101) - (1 + 2 + 3 + 4 + ... + 99 + 100)
Đặt B = 1.2 + 2.3 + 3.4 + ... + 99.100 + 100.101
=> 3B = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3 + 100.101.3
= 1.2.3 + 2.3(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98) + 100.101.(102 - 99)
= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100 + 100.101.102 - 99.100.101
= 100.101.102
= 1 030 200
=> B = 343 400
Khi đó : C = B - (1 + 2 + 3 + 4 + ... + 99 + 100)
= 343 400 - [(100 - 1) : 1 + 1] . (100 + 1) : 2
= 343 400 - 100 . 101 : 2
= 343 400 + 5050
= 348 450
Vậy C = 348 500
a) Ta có : A = 3 + 32 + 33 + 34 + ... + 399 + 3100
= (3 + 32) + (33 + 34) + ... + (399 + 3100)
= (3 + 32) + 32.(3 + 32) + .... + 398.(3 + 32)
= 12 + 32.12 + .... + 398.12
= 12.(1 + 32 + ... + 398) (1)
= 3.4.(1 + 32 + ... + 398) \(⋮\) 4
=> \(A⋮4\)
Từ (1) \(\Rightarrow A⋮12\)
b) B = 1 x 2 + 2 x 3 + 3 x 4 + ... + 99 x 100
3B = 1 x 2 x 3 + 2 x 3 x 3 + 3 x 4 x 3 + ... + 99 x 100 x 3
= 1 x 2 x 3 + 2 x 3 x (4 - 1) + 3 x 4 x (5 - 2) + .... + 99 x 100 x (101 - 98)
= 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5 - 2 x 3 x 4 + .... + 99 x 100 x 101 - 98 x 99 x 100
= 99 x 100 x 101 = 999 900
=> B = 333 300
c) Ta có : C = 12 + 22 + 32 + ... + 992 + 1002
= 1.1 + 2.2 + 3.3 + ... + 99.99 + 100.100
= 1.(2 - 1) + 2.(3 - 1) + 3.(4 - 1) + .... + 99.(100 - 1) + 100.(101 - 1)
= 1.2 - 1 + 2.3 - 2 + 3.4 - 3 + ... + 99.100 - 99 + 100.101 - 100
= (1.2 + 2.3 + 3.4 + ... + 99.100 + 100.101) - (1 + 2 + 3 + 4 + ... + 99 + 100)
Đặt B = 1.2 + 2.3 + 3.4 + ... + 99.100 + 100.101
=> 3B = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3 + 100.101.3
= 1.2.3 + 2.3(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98) + 100.101.(102 - 99)
= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100 + 100.101.102 - 99.100.101
= 100.101.102
= 1 030 200
=> B = 343 400
Khi đó : C = B - (1 + 2 + 3 + 4 + ... + 99 + 100)
= 343 400 - [(100 - 1) : 1 + 1] . (100 + 1) : 2
= 343 400 - 100 . 101 : 2
= 343 400 + 5050
= 348 450
Vậy C = 348 500