K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
29 tháng 11 2016
b, n = 3 hoặc 7
c, n = 4 hoặc 5 hoặc 7
d, n = 6 hoặc 7 hoặc 8 hoặc 11
d, n=2 hoặc 3 hoặc 5
26 tháng 2 2017
Bài 1:
b) Ta có:
\(16^5=2^{20}\)
\(\Rightarrow B=16^5+2^{15}=2^{20}+2^{15}\)
\(\Rightarrow B=2^{15}.2^5+2^{15}\)
\(\Rightarrow B=2^{15}\left(2^5+1\right)\)
\(\Rightarrow B=2^{15}.33\)
\(\Rightarrow B⋮33\) (Đpcm)
c) \(C=5+5^2+5^3+5^4+...+5^{100}\)
\(\Rightarrow C=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{99}+5^{100}\right)\)
\(\Rightarrow C=1\left(5+5^2\right)+5^2\left(5+5^2\right)+...+5^{98}\left(5+5^2\right)\)
\(\Rightarrow\left(1+5^2+...+5^{98}\right)\left(5+5^2\right)\)
\(\Rightarrow C=Q.30\)
\(\Rightarrow C⋮30\) (Đpcm)
26 tháng 2 2017
Bài 1 : a, \(A=1+3+3^2+...+3^{118}+3^{119}\)
\(A=\left(1+3+3^2+3^3\right)+...+\left(3^{116}+3^{117}+3^{118}+3^{119}\right)\)
\(A=\left(1+3+3^2+3^3\right)+...+3^{116}\left(1+3+3^2+3^3\right)\)
\(A=1.30+...+3^{116}.30=\left(1+...+3^{116}\right).30⋮3\)
Vậy \(A⋮3\)
b, \(B=16^5+2^{15}=\left(2.8\right)^5+2^{15}\)
\(=2^5.8^5+2^{15}=2^5.\left(2^3\right)^5+2^{15}\)
\(=2^5.2^{15}+2^{15}.1=2^{15}\left(32+1\right)=2^{15}.33⋮33\)
Vậy \(B⋮33\)
c, Tương tự câu a nhưng nhóm 2 số
Bài 2 : a, \(n+2⋮n-1\) ; Mà : \(n-1⋮n-1\)
\(\Rightarrow\left(n+2\right)-\left(n-1\right)⋮n-1\)
\(\Rightarrow n+2-n+1⋮n-1\Rightarrow3⋮n-1\)
\(\Rightarrow n-1\in\left\{1;3\right\}\Rightarrow n\in\left\{2;4\right\}\)
Vậy \(n\in\left\{2;4\right\}\) thỏa mãn đề bài
b, \(2n+7⋮n+1\)
Mà : \(n+1⋮n+1\Rightarrow2\left(n+1\right)⋮n+1\Rightarrow2n+2⋮n+1\)
\(\Rightarrow\left(2n+7\right)-\left(2n+2\right)⋮n+1\)
\(\Rightarrow2n+7-2n-2⋮n+1\Rightarrow5⋮n+1\)
\(\Rightarrow n+1\in\left\{1;5\right\}\Rightarrow n\in\left\{0;4\right\}\)
Vậy \(n\in\left\{0;4\right\}\) thỏa mãn đề bài
c, tương tự phần b
d, Vì : \(4n+3⋮2n+6\)
Mà : \(2n+6⋮2n+6\Rightarrow2\left(2n+6\right)⋮2n+6\Rightarrow4n+12⋮2n+6\)
\(\Rightarrow\left(4n+12\right)-\left(4n+3\right)⋮2n+6\)
\(\Rightarrow4n+12-4n-3⋮2n+6\Rightarrow9⋮2n+6\)
\(\Rightarrow2n+6\in\left\{1;2;9\right\}\Rightarrow2n=3\Rightarrow n\in\varnothing\)
Vậy \(n\in\varnothing\)
5 tháng 5 2017
e, D = 512+1 /513+ 1 < 1 => 512+1/ 513+1 < 512+1+4/ 513+1+4
= 512+5/ 513+5 = 5. (511+1) / 5. (512+1) = 511+1 / 512+1= E
Vậy D < E
22 tháng 5 2022
a: \(n+6⋮n\)
nên \(n\inƯ\left(6\right)\)
hay \(n\in\left\{1;2;3;6\right\}\)
b: \(38-3n⋮n\)
nên \(n\inƯ\left(38\right)\)
hay \(n\in\left\{1;2;19;38\right\}\)
c: \(n+5⋮n+1\)
\(\Leftrightarrow n+1+4⋮n+1\)
\(\Leftrightarrow n+1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(n\in\left\{0;1;3\right\}\)
d: \(\Leftrightarrow n+1\in\left\{1;2;4;7;14;28\right\}\)
hay \(n\in\left\{0;1;3;6;13;27\right\}\)
8 tháng 5 2017
Ta có :
\(A=\dfrac{1}{5^2}+\dfrac{2}{5^3}+\dfrac{3}{5^4}+.............+\dfrac{n}{5^{n+1}}+.....+\dfrac{11}{5^{12}}\)
\(\Rightarrow5A=\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{3^3}+........+\dfrac{n}{5^n}+..........+\dfrac{11}{5^{11}}\)
\(\Rightarrow5A-A=\left(\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+.....+\dfrac{n}{5^n}+....+\dfrac{11}{5^{11}}\right)-\left(\dfrac{1}{5^2}+\dfrac{2}{5^3}+.....+\dfrac{n}{5^{n+1}}+........+\dfrac{11}{5^{12}}\right)\)\(\Rightarrow4A=\dfrac{1}{5}+\dfrac{1}{5^2}+........+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\)
\(\Rightarrow20A=1+\dfrac{1}{5}+.........+\dfrac{1}{5^{10}}-\dfrac{11}{5^{11}}\)
\(\Rightarrow20A-4A=\left(1+\dfrac{1}{5}+.......+\dfrac{1}{5^{10}}-\dfrac{11}{5^{11}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+........+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\right)\)\(\Rightarrow16A=1-\dfrac{12}{5^{11}}+\dfrac{11}{5^{12}}< 1\)
\(\Rightarrow A< \dfrac{1}{16}\rightarrowđpcm\)
Ta có :
Mà : Ư(4) =
*TH1 :
* TH2:
* TH3:
Vậy :
\(\left(n+5\right)⋮\left(n-1\right)\)
\(\Rightarrow\)\(\left(n-1+6\right)⋮\left(n-1\right)\)
\(\Rightarrow\)\(\left(n-1\right)+6⋮\left(n-1\right)\)
Vì \(\left(n-1\right)⋮\left(n-1\right)\)nên để \(\left(n+5\right)⋮\left(n-1\right)\)thì \(6⋮\left(n-1\right)\)
\(\Rightarrow\)\(\left(n-1\right)\inƯ\left(6\right)\)
\(\Rightarrow\)\(\left(n-1\right)\in\left\{1,-1,2,-2,3,-3,6,-6\right\}\)
\(\Rightarrow\)\(n\in\left\{2,0,3,-1,4,-2,7,-5\right\}\)
Vậy \(n\in\left\{2,0,3,-1,4,-2,7,-5\right\}\)