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b) n mũ 2 + 2006 là hợp số
hai câu còn lại ko bt
Hok tốt
^_^
b1
ta có : n+4 = (n+1)+3
=>n+1+3 chia hết cho n+1
vì n+1 chia hết cho n+1
=>3 chia hết cho n+1
=> n+1 chia hết cho 3
=> n+1 thuộc Ư 3 =[1;3]
=> n+1=1 n+1=3
n =1-1 n =3-1
n =0 n =2
vậy n thuộc [0;2]
Mình gõ câu a bị lỗi nha , thực chất câu a là
a) Tìm các số tự nhiên x, y biết : 2xy + x + 2y = 13
a)Bạn làm nha vì bài này dễ rồi
b)+)Ta có:A=1.2+2.3+3.4+..................+99.100
=>3A=1.2.3+2.3.3+3.4.3+.................+99.100.3
=>3A=1.2.3+2.3.(4-1)+3.4.(5-2)+................+99.100.(101-98)
=>3A=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...................-98.99.100+99.100.101
=>3A=99.100.101
=>A=\(\frac{99.100.101}{3}=333300\)
+)Ta lại có:B=12+22+32+..................+992
=>B=1.1+2.2+3.3+............+99.99
=>B=1.(2-1)+2.(3-1)+3.(4-1)+..........+99.(100-1)
=>B=1.2-1+2.3-2+3.4-3+........................+99.100-99
=>B=(1.2+2.3+3.4+............+99.100)-(1+2+3+..............+99)
Đặt N=1.2+2.3+3.4+....................+99.100
=>3N=1.2.3+2.3.3+3.4.3+.................+99.100.3
=>3N=1.2.3+2.3.(4-1)+3.4.(5-2)+................+99.100.(101-98)
=>3N=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...................-98.99.100+99.100.101
=>3N=99.100.101
=>N=\(\frac{99.100.101}{3}=333300\)
Đặt M=1+2+3+..............+99(có 99 số hạng)
=>M=\(\frac{\left(1+99\right).99}{2}=4950\)
+)Ta thấy A-B=333300-(333300-4950)
=>A-B=333300-333300+4950
=>A-B=4950\(⋮\)50
Vậy A-B\(⋮\)50
Chúc bn học tốt
\(3+3^2+3^3+...+3^{2012}\)
\(=\left(3+3^2+3^3+3^4\right)+...+\left(3^{2009}+3^{2010}+3^{2011}+3^{2012}\right)\)
\(=3\left(1+3+3^2+3^3\right)+...+3^{2009}\left(1+3+3^2+3^3\right)\)
\(=40\left(3+...+3^{2009}\right)⋮40\)
a)xét 2A =2+2^2+2^3+.....+2^2019
-A=1+2+2^2+...+2^2018
A=(2^2019)-1 <2^2019
b)theo câu a ta có A+1=2^2019-1+1=2^2019=2^(x+1)
2019=x+1 =>x=2018
Bài 1:
a) Ta có: \(\left(2x-1\right)^{20}=\left(2x-1\right)^{18}\)
\(\Leftrightarrow\left(2x-1\right)^{20}-\left(2x-1\right)^{18}=0\)
\(\Leftrightarrow\left(2x-1\right)^{18}\left[\left(2x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left(2x-1\right)^{18}\cdot\left(2x-2\right)\cdot2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
b) Ta có: \(\left(2x-3\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
c) Ta có: \(\left(x-5\right)^2=\left(1-3x\right)^2\)
\(\Leftrightarrow\left(x-5\right)^2-\left(3x-1\right)^2=0\)
\(\Leftrightarrow\left(x-5-3x+1\right)\left(x-5+3x-1\right)=0\)
\(\Leftrightarrow\left(-2x-4\right)\left(4x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{2}\end{matrix}\right.\)
Bài 2:
a) \(15^{20}-15^{19}=15^{19}\left(15-1\right)=15^{19}\cdot14⋮14\)
b) \(3^{20}+3^{21}+3^{22}=3^{20}\left(1+3+3^2\right)=3^{20}\cdot13⋮13\)
c) \(3+3^2+3^3+...+3^{2007}\)
\(=3\left(1+3+3^2\right)+...+3^{2005}\left(1+3+3^2\right)\)
\(=13\left(3+...+3^{2005}\right)⋮13\)
Ta có :
A= 32+33+34+35+...+350+351
A= (32+33)+(34+35)+...+(350+351)
A= 1(32+33)+32(32+33)+...+348(32+33)
A= 1.36 + 32.36+...+348.36
A= 36(1+32+...+348) \(⋮36\)
Vì A \(⋮36\) mà 36 \(⋮12\)=> A \(⋮12\)
A = (3^2+3^3)+(3^4+3^5)+....+(3^50+3^51)
= 3.(3+3^2)+3^3.(3+3^2)+....+3^49.(3+3^2)
= 3.12 + 3^3.12 + .... +3^49.12
= 12.(3+3^3+....+3^49) chia hết cho 12 (ĐPCM)
Bài 1:
a,\(A=3+3^2+3^3+...+3^{2010}\)
\(=\left(3+3^2+3^3+3^4\right)+....+\left(3^{2007}+3^{2008}+3^{2009}+3^{2010}\right)\)
\(=3\left(1+3+3^2+3^3\right)+....+3^{2007}\left(1+3+3^2+3^3\right)\)
\(=3.40+...+3^{2007}.40\)
\(=40\left(3+3^5+...+3^{2007}\right)⋮40\)
Vì A chia hết cho 40 nên chữ số tận cùng của A là 0
b,\(A=3+3^2+3^3+...+3^{2010}\)
\(3A=3^2+3^3+...+3^{2011}\)
\(3A-A=\left(3^2+3^3+...+3^{2011}\right)-\left(3+3^2+3^3+...+3^{2010}\right)\)
\(2A=3^{2011}-3\)
\(2A+3=3^{2011}\)
Vậy 2A+3 là 1 lũy thừa của 3
a: \(A=3+3^2+3^3+...+3^{2024}\)
\(=\left(3+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{2021}+3^{2022}+3^{2023}+3^{2024}\right)\)
\(=\left(3+3^2+3^3+3^4\right)+3^4\left(3+3^2+3^3+3^4\right)+...+3^{2020}\left(3+3^2+3^3+3^4\right)\)
\(=120\left(1+3^4+...+3^{2020}\right)⋮120\)
b: x+10y+2xy+1=0
=>x+2xy+10y+1=0
=>x(2y+1)+10y+5-4=0
=>x(2y+1)+5(2y+1)=4
=>(2y+1)*(x+5)=4
mà 2y+1 lẻ(do y nguyên)
nên \(\left(x+5;2y+1\right)\in\left\{\left(4;1\right);\left(-4;-1\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(-1;0\right);\left(-9;-1\right)\right\}\)
a)A=(3^1+3^2+3^3+3^4)+(3^5+3^6+3^7+3^8)+...+(3^2021+3^2022+3^2023+3^2024)
A=120+3^4(3+3^2+3^3+3^4)+..+3^2020(3+3^2+3^3+3^4)
A=120+3^4.120+...+3^2020.120
A=120(1+3^4+...+3^2020)⋮120