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a) \(A=2+2^2+2^3+\dots+2^{60}\)
\(2A=2^2+2^3+2^4+\dots+2^{61}\)
\(2A-A=\left(2^2+2^3+2^4+\dots+2^{61}\right)-\left(2+2^2+2^3+\dots+2^{60}\right)\)
\(A=2^{61}-2\)
Vậy: \(A=2^{61}-2\).
b)
+) \(A=2+2^2+2^3+\dots+2^{60}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+\left(2^5+2^6\right)+\dots+\left(2^{59}+2^{60}\right)\)
\(=2\cdot\left(1+2\right)+2^3\cdot\left(1+2\right)+2^5\cdot\left(1+2\right)+\dots+2^{59}\cdot\left(1+2\right)\)
\(=2\cdot3+2^3\cdot3+2^5\cdot3+\dots+2^{59}\cdot3\)
\(=3\cdot\left(2+2^3+2^5+\dots+2^{59}\right)\)
Vì \(3\cdot\left(2+2^3+2^5+\dots+2^{59}\right)⋮3\) nên \(A⋮3\)
+) \(A=2+2^2+2^3+\dots+2^{60}\)
\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+\left(2^9+2^{10}+2^{11}+2^{12}\right)+\dots+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(=2\cdot\left(1+2+2^2+2^3\right)+2^5\cdot\left(1+2+2^2+2^3\right)+2^9\cdot\left(1+2+2^2+2^3\right)+\dots+2^{57}\cdot\left(1+2+2^2+2^3\right)\)
\(=2\cdot15+2^5\cdot15+2^9\cdot15+\dots+2^{57}\cdot15\)
\(=15\cdot\left(2+2^5+2^9+\dots+2^{57}\right)\)
Vì \(15⋮5\) nên \(15\cdot\left(2+2^5+2^9+\dots+2^{57}\right)⋮5\)
hay \(A\vdots5\)
+) \(A=2+2^2+2^3+\dots+2^{60}\)
\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+\left(2^7+2^8+2^9\right)+\dots+\left(2^{58}+2^{59}+2^{60}\right)\)
\(=2\cdot\left(1+2+2^2\right)+2^4\cdot\left(1+2+2^2\right)+2^7\cdot\left(1+2+2^2\right)+\dots+2^{58}\cdot\left(1+2+2^2\right)\)
\(=2\cdot7+2^4\cdot7+2^7\cdot7+\dots+2^{58}\cdot7\)
\(=7\cdot\left(2+2^4+2^7+\dots+2^{58}\right)\)
Vì \(7\cdot\left(2+2^4+2^7+\dots+2^{58}\right)⋮7\) nên \(A⋮7\)
$Toru$
a) �=2+22+23+⋯+260A=2+22+23+⋯+260
2�=22+23+24+⋯+2612A=22+23+24+⋯+261
2�−�=(22+23+24+⋯+261)−(2+22+23+⋯+260)2A−A=(22+23+24+⋯+261)−(2+22+23+⋯+260)
�=261−2A=261−2
Vậy: �=261−2A=261−2.
b)
+) �=2+22+23+⋯+260A=2+22+23+⋯+260
=(2+22)+(23+24)+(25+26)+⋯+(259+260)=(2+22)+(23+24)+(25+26)+⋯+(259+260)
=2⋅(1+2)+23⋅(1+2)+25⋅(1+2)+⋯+259⋅(1+2)=2⋅(1+2)+23⋅(1+2)+25⋅(1+2)+⋯+259⋅(1+2)
=2⋅3+23⋅3+25⋅3+⋯+259⋅3=2⋅3+23⋅3+25⋅3+⋯+259⋅3
=3⋅(2+23+25+⋯+259)=3⋅(2+23+25+⋯+259)
Vì 3⋅(2+23+25+⋯+259)⋮33⋅(2+23+25+⋯+259)⋮3 nên �⋮3A⋮3
+) �=2+22+23+⋯+260A=2+22+23+⋯+260
=(2+22+23+24)+(25+26+27+28)+(29+210+211+212)+⋯+(257+258+259+260)=(2+22+23+24)+(25+26+27+28)+(29+210+211+212)+⋯+(257+258+259+260)
=2⋅(1+2+22+23)+25⋅(1+2+22+23)+29⋅(1+2+22+23)+⋯+257⋅(1+2+22+23)=2⋅(1+2+22+23)+25⋅(1+2+22+23)+29⋅(1+2+22+23)+⋯+257⋅(1+2+22+23)
=2⋅15+25⋅15+29⋅15+⋯+257⋅15=2⋅15+25⋅15+29⋅15+⋯+257⋅15
=15⋅(2+25+29+⋯+257)=15⋅(2+25+29+⋯+257)
Vì 15⋮515⋮5 nên 15⋅(2+25+29+⋯+257)⋮515⋅(2+25+29+⋯+257)⋮5
hay �⋮5A⋮5
+) �=2+22+23+⋯+260A=2+22+23+⋯+260
=(2+22+23)+(24+25+26)+(27+28+29)+⋯+(258+259+260)=(2+22+23)+(24+25+26)+(27+28+29)+⋯+(258+259+260)
=2⋅(1+2+22)+24⋅(1+2+22)+27⋅(1+2+22)+⋯+258⋅(1+2+22)=2⋅(1+2+22)+24⋅(1+2+22)+27⋅(1+2+22)+⋯+258⋅(1+2+22)
=2⋅7+24⋅7+27⋅7+⋯+258⋅7=2⋅7+24⋅7+27⋅7+⋯+258⋅7
=7⋅(2+24+27+⋯+258)=7⋅(2+24+27+⋯+258)
Vì 7⋅(2+24+27+⋯+258)⋮77⋅(2+24+27+⋯+258)⋮7 nên �⋮7A⋮7
Gọi x (cuốn) là số sách cần tìm (x ∈ ℕ* và 99 < x < 1000)
Do khi xếp thành từng bó 18 cuốn, 22 cuốn, 24 cuốn đều thừa ra 5 cuốn nên x - 5 ∈ BC(18; 22; 24)
Ta có:
18 = 2.3²
22 = 2.11
24 = 2³.3
⇒ BCNN(18; 22; 24) = 2³.3².11 = 792
⇒ x - 5 ∈ BC(18; 22; 24) = B(792) = {0; 792; 1584; ...}
⇒ x ∈ {5; 797; 1589; ...}
Mà 99 < x < 1000
⇒ x = 797
Vậy số cuốn sách cần tìm là 797 cuốn
Theo thầy thì vào tất cả các bài trong HKI nhé em
Ta có: n2+4=(n-3).n+3n+4
Vì n2+4⋮n-3 và (n-3).n⋮n-3 nên 3n+4⋮n-3
Lại có: 3n+4=(n-3).3+13
Vì 3n+4⋮n-3 và (n-3).3⋮n-3 nên 13⋮n-3
⇒ n-3ϵ{1;-1;13;-13}
⇒ nϵ{4;2;16;-10}
Vậy ...
Lời giải:
a. Với $x,y$ nguyên thì $x-2, 2y+1$ nguyên.
Mà $(x-2)(2y+1)=8$ nên $2y+1$ là ước của $8$
$2y+1$ lẻ nên $2y+1=1$ hoặc $2y+1=-1$
Nếu $2y+1=1\Rightarrow x-2=8$
$\Rightarrow y=0; x=10$
Nếu $2y+1=-1\Rightarrow x-2=-8$
$\Rightarrow y=-1; x=-6$
b.
$8-x, 4y+1$ là số nguyên. Mà $(8-x)(4y+1)=20$ nên $4y+1$ là ước của $20$.
Mà $4y+1$ chia $4$ dư $1$ nên $4y+1\in \left\{1; 5\right\}$
Nếu $4y+1=1$ thì $8-x=20$
$\Rightarrow y=0; x=-12$
Nếu $4y+1=5$ thì $8-x=4$
$\Rightarrow y=1; x=4$
(2\(x\) - 1).(5 - y) = 24
24 = 23.3
Ư(24) = {-24; -12; - 8; -6; -4; -3; -2; -1; 1; 2; 3; 4; 6; 8; 12; 24}
Lập bảng ta có:
| 2\(x\) - 1 | -24 | -12 | -8 | -6 | -4 | -3 | -2 | -1 | 1 | 2 | 3 | 4 | 6 | 8 | 12 | 24 |
| \(x\) | -\(\dfrac{23}{5}\) | -\(\dfrac{11}{2}\) | -\(\dfrac{7}{2}\) | -\(\dfrac{5}{2}\) | \(-\dfrac{3}{2}\) | -1 | -\(\dfrac{1}{2}\) | 0 | 1 | \(\dfrac{3}{2}\) | 2 | \(\dfrac{5}{2}\) | \(\dfrac{7}{2}\) | \(\dfrac{9}{2}\) | \(\dfrac{13}{2}\) | \(\dfrac{25}{2}\) |
| 5 - y | -1 | -2 | -3 | -4 | -6 | -8 | -12 | -24 | 24 | 12 | 8 | 6 | 4 | 3 | 2 | 1 |
| y | 6 | 7 | 8 | 9 | 11 | 13 | 17 | 29 | -19 | -7 | -3 | -1 | 1 | 2 | 3 | 4 |
Vậy (\(x\); y) = (-1; 13); (0; 29); (1; -19); (2; -3);
6) 31 - (17 + x) = 18
17 + x = 31 - 18
17 + x = 13
x = 13 - 17
x = -4
7) (-x + 31) - 39 = -69 + 11
31 - x - 39 = -58
-8 - x = -58
x = -8 - (-58)
x = 50
8) x - (18 - 13) = 5 - (24 - 27)
x - 5 = 5 - (-3)
x - 5 = 8
x = 8 + 5
x = 13
9) 315 - (x + 315) = 43 + (9 - 21)
315 - x - 315 = 43 - 12
-x = 31
x = -31
10) 25 - (25 - x) = 12 + (42 - 65)
25 - 25 + x = 12 - 23
x = -11