A=20+21+...+211
A chia hết cho 3;7;15
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Giải:
a) \(M=21^9+21^8+21^7+...+21+1\)
Do \(21^n\) luôn có tận cùng là 1
\(\Rightarrow M=21^9+21^8+21^7+...+21+1\)
Tân cùng của M là:
\(1+1+1+1+1+1+1+1+1+1=10\) tận cùng là 0
\(\Rightarrow M⋮10\)
\(\Leftrightarrow M⋮2;5\)
b) \(N=6+6^2+6^3+...+6^{2020}\)
\(N=6.\left(1+6\right)+6^3.\left(1+6\right)+...+6^{2019}.\left(1+6\right)\)
\(N=6.7+6^3.7+...+6^{2019}.7\)
\(N=7.\left(6+6^3+...+6^{2019}\right)⋮7\)
\(\Rightarrow N⋮7\)
Ta thấy: \(N=6+6^2+6^3+...+6^{2020}⋮6\)
Mà \(6⋮̸9\)
\(\Rightarrow N⋮̸9\)
c) \(P=4+4^2+4^3+...+4^{23}+4^{24}\)
\(P=1.\left(4+4^2\right)+4^2.\left(4+4^2\right)+...+4^{20}.\left(4+4^2\right)+4^{22}.\left(4+4^2\right)\)
\(P=1.20+4^2.20+...+4^{20}.20+4^{22}.20\)
\(P=20.\left(1+4^2+...+4^{20}+4^{22}\right)⋮20\)
\(\Rightarrow P⋮20\)
\(P=4+4^2+4^3+...+4^{23}+4^{24}\)
\(P=4.\left(1+4+4^2\right)+...+4^{22}.\left(1+4+4^2\right)\)
\(P=4.21+...+4^{22}.21\)
\(P=21.\left(4+...+4^{22}\right)⋮21\)
\(\Rightarrow P⋮21\)
d) \(Q=6+6^2+6^3+...+6^{99}\)
\(Q=6.\left(1+6+6^2\right)+...+6^{97}.\left(1+6+6^2\right)\)
\(Q=6.43+...+6^{97}.43\)
\(Q=43.\left(6+...+6^{97}\right)⋮43\)
\(\Rightarrow Q⋮43\)
Chúc bạn học tốt!
Lời giải:
$A=(4+4^2)+(4^3+4^4)+....+(4^{23}+4^{24})$
$=(4+4^2)+4^2(4+4^2)+....+4^{22}(4+4^2)$
$=(4+4^2)(1+4^2+...+4^{22})$
$=20(1+4^2+...+4^{22})\vdots 20$
----------------------------
$A=(4+4^2+4^3)+(4^4+4^5+4^6)+....+(4^{22}+4^{23}+4^{24})$
$=4(1+4+4^2)+4^4(1+4+4^2)+....+4^{22}(1+4+4^2)$
$=(1+4+4^2)(4+4^4+...+4^{22})$
$=21(4+4^4+....+4^{22})\vdots 21$
----------------------
Vậy $A\vdots 20; A\vdots 21$. Mà $(20,21)=1$ nên $A\vdots (20.21)$ hay $A\vdots 420$
A = (4 + 4^2 + 4^3 + 4^4 + 4^5 + 4^6) + (4^7 + 4^8 + 4^9 + 4^10 + 4^11 + 4^12) + (4^13 + 4^14 + 4^15 + 4^16 + 4^17 + 4^18) + (4^19 + 4^20 + 4^21 + 4^22 + 4^23 + 4^24)
A = (4 + 4^2 + 4^3 + 4^4 + 4^5 + 4^6) + 4^6(4 + 4^2 + 4^3 + 4^4 + 4^5 + 4^6) + 4^12(4 + 4^2 + 4^3 + 4^4 + 4^5 + 4^6) + 4^18(4 + 4^2 + 4^3 + 4^4 + 4^5 + 4^6)
A = (4 + 4^2 + 4^3 + 4^4 + 4^5 + 4^6).(1+4^6+4^12+4^18)
A = 5460.(1+4^6+4^12+4^18)
A = 420 . 13(1+4^6+4^12+4^18) => A chia hết cho 420
A = 20.21.13(1+4^6+4^12+4^18) => A chia hết cho 20 ; 21
A = \(4+4^2+4^3+.....+4^{23}+4^{24}\)
= \(4\left(1+4+4^2\right)+.....+4^{22}+\left(1+4+4^2\right)\)
= \(4.21+.....+4^{22}.21\)
= \(21\left(4+...+4^{22}\right)⋮21\)
Vậy A chia hết cho 21
Ai k mik mik k lại nha
Lâu r chị k nhớ lắm nhé
CM A chia hết cho 20
A = 4(1+4+4^2+...+4^23) chia hết cho 4 (1)
A = (4+4^2) + (4^3+4^4) + ...+ (4^23+4^24)
= 4(1+4) + 4^3(1+4) +...+4^23(1+4)
= (1+4)(4+4^3+4^5+...+4^23)
=5.(4+4^3+4^5+...+4^23) chia hết cho 5 (2)
Mà UCLN(4,5)=1 (3)
Vậy A chia hết cho 4.5 =20
CM A chia hết cho 21
A = (4+4^2+4^3)+(4^4+4^5+4^6)+...+(4^22+4^23+4^24)
= 4(1+4+4^2) +4^4(1+4+4^2)+...+4^22(1+4+4^2)
= (1+4+4^2)(4+4^4+...+4^22)
= 21(4+4^4+...+4^22) chia hết cho 21
Vậy A chia hết cho 24.
Chúc e học giỏi!
1) ta có A= 4+4^2 +4^3 +4^4 +...+4^120 =( 4+ 4^2 )+ (4^3+4^4) +...+ (4^119+4^120)
=4.(1+4) +4^3.(1+4) +...+4^119.(1+4) = (1+4).(4+4^3+...+4^119) =5 .(4+4^3+..+4^119)
mà 4+4^3+4^119 chia hết cho 4 , UCLN(4,5)=1 =>5.(4+4^3+...+4^119) chia het cho 20 => A chia het cho 20
2) ta coA= 4+4^2+4^3 +...+4^120 = (4+4^2+4^3) +...+ (4^118+4^119+4^120)
=4.(1+4+4^2)+...+4^118.(1+4+4^2) = 21.( 4+..+4^118) chia het cho 21 => A chia het cho 21
do A chia het cho 20, 21 mà UCLN(20,21) =1 nên A chia hết cho 20 .21 => A chia hết cho 420
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) \(3+3^2+3^3+...+3^{2007}\)
\(=3\left(1+3+3^2\right)+...+3^{2005}\left(1+3+3^2\right)\)
\(=13\left(3+...+3^{2005}\right)⋮13\)
4 + 42 + 43 + 44 + ... + 423 + 424
= 4x(1+4) + 42x4x(1+4) + ... + 422x4x(1+4)
= 20 + 42x20 + ... + 422x20
= 20x(1+42+...+422)
Suy ra: A chia hết cho 20
4 + 42 + 43 + 44 + ... + 423 + 424
= (4 + 42 + 43) + ... + (422 + 423 + 424)
= 4x(1+4+42) + ... + 422x(1+4+42)
= 4x21 + ... + 422x21
= (4+...+422)x21
Suy ra: A chia hết cho 21
Vì A chia hết cho 21 , A chia hết cho 20
Suy ra: A chia hết cho 21x20=420
1 +1 = 0 + 2