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Ta có: \(C=3-3^2+3^3-3^4+\cdots+3^{23}-3^{24}\)
\(=3\left(1-3+3^2-3^3+\cdots+3^{22}-3^{23}\right)\) ⋮3
Ta có: \(C=3-3^2+3^3-3^4+\cdots+3^{23}-3^{24}\)
\(=3\left(1-3\right)+3^3\left(1-3\right)+\cdots+3^{23}\left(1-3\right)\)
\(=3\cdot\left(-2\right)+3^3\cdot\left(-2\right)+\cdots+3^{23}\left(-2\right)\)
\(=-2\left(3+3^3+\cdots+3^{23}\right)\)
\(=-2\left\lbrack\left(3+3^3\right)+\left(3^5+3^7\right)+\cdots+\left(3^{21}+3^{23}\right)\right\rbrack\)
\(=-2\cdot\left\lbrack3\cdot\left(1+3^2\right)+3^5\left(1+3^2\right)+\cdots+3^{21}\left(1+3^2\right)\right\rbrack\)
\(=-2\cdot10\cdot\left(3+3^5+\cdots+3^{21}\right)=-20\cdot\left(3+3^5+\cdots+3^{21}\right)\) ⋮20
Ta có: \(C=3-3^2+3^3-3^4+\cdots+3^{23}-3^{24}\)
\(=\left(3-3^2+3^3\right)-\left(3^4-3^5+3^6\right)+\cdots-\left(3^{22}-3^{23}+3^{24}\right)\)
\(=3\left(1-3+3^2\right)-3^4\left(1-3+3^2\right)+\cdots+3^{22}\left(1-3+3^2\right)\)
\(=7\cdot\left(3-3^4+\cdots-3^{22}\right)\) ⋮7
Ta có: C⋮20
C⋮7
C⋮3
mà ƯCLN(20;3;7)=1
nên C⋮20*3*7
=>C⋮420
1)Tính:
a)\(4^2\cdot2=\left(2^2\right)^2\cdot2=2^4\cdot2=2^5=32\)
b)\(36^2:6^2=\left(36:6\right)^2=6^2=48\)
c)\(\left(\frac{2}{5}\right)^{10}:\left(\frac{4}{25}\right)^2=\left(\frac{2}{5}\right)^{10}\cdot\left(\frac{25}{4}\right)^2=\)\(\left(1\right)^{10}\cdot\left(\frac{5}{2}\right)^2=1\cdot\frac{5^2}{2^2}=1\cdot\frac{25}{4}=\frac{25}{4}\)
a
\(4^2.2=16.2=32\)
b\(36^2:6^2=36.36:6.6=36.36:36=36\)
c
Ta có: \(C=3-3^2+3^3-3^4+\cdots+3^{23}-3^{24}\)
\(=3\left(1-3+3^2-3^3+\cdots+3^{22}-3^{23}\right)\) ⋮3
Ta có: \(C=3-3^2+3^3-3^4+\cdots+3^{23}-3^{24}\)
\(=3\left(1-3\right)+3^3\left(1-3\right)+\cdots+3^{23}\left(1-3\right)\)
\(=3\cdot\left(-2\right)+3^3\cdot\left(-2\right)+\cdots+3^{23}\left(-2\right)\)
\(=-2\left(3+3^3+\cdots+3^{23}\right)\)
\(=-2\left\lbrack\left(3+3^3\right)+\left(3^5+3^7\right)+\cdots+\left(3^{21}+3^{23}\right)\right\rbrack\)
\(=-2\cdot\left\lbrack3\cdot\left(1+3^2\right)+3^5\left(1+3^2\right)+\cdots+3^{21}\left(1+3^2\right)\right\rbrack\)
\(=-2\cdot10\cdot\left(3+3^5+\cdots+3^{21}\right)=-20\cdot\left(3+3^5+\cdots+3^{21}\right)\) ⋮20
Ta có: \(C=3-3^2+3^3-3^4+\cdots+3^{23}-3^{24}\)
\(=\left(3-3^2+3^3\right)-\left(3^4-3^5+3^6\right)+\cdots-\left(3^{22}-3^{23}+3^{24}\right)\)
\(=3\left(1-3+3^2\right)-3^4\left(1-3+3^2\right)+\cdots+3^{22}\left(1-3+3^2\right)\)
\(=7\cdot\left(3-3^4+\cdots-3^{22}\right)\) ⋮7
Ta có: C⋮20
C⋮7
C⋮3
mà ƯCLN(20;3;7)=1
nên C⋮20*3*7
=>C⋮420
Bạn phân tích nhu mình vừa nãy thì sẽ có \(a=\frac{10^{2n}-1}{9}\) \(b=\frac{10^{n+1}-1}{9},c=\frac{6\left(10^n-1\right)}{9}\)
cộng tất cả vào ta sẽ có a+b+c+8 ( 8 =72/9) và bằng
\(\frac{10^{2n}-1+10^{n+1}-1+6\left(10^n-1\right)+72}{9}\)
phân tích 10^2n = (10^n)^2
10^(n+1) = 10^n.10 và 6(10^n-1) thành 6.10^n-6 và cộng 72-1-1=70, ta được
\(\frac{\left(10^n\right)^2+10^n.10+6.10^n-6+70}{9}\)
=\(\frac{\left(10^n\right)^2+10^n.16+64}{9}\)
=\(\frac{\left(10^n+8\right)^2}{3^2}\)
=\(\left(\frac{10^n+8}{3}\right)^2\)
vì 10^n +8 có dạng 10000..08 nên chia hết cho 3 => a+b+c+8 là số chính phương
92:33=(32)2:33=34:33=34–3=3
52.252= 52.(52)2=52.54=52+4=56
a)92 : 33 = (32)2 : 33 = 34 : 33 = 3.
b) 52 . 252 = 52 . (52)2 = 52 . 54 = 56.
c) \(\left(\frac{1}{3}\right)^2\) . \(\left(\frac{1}{9.3}\right)^2\) = \(\frac{1^2}{3^2}\). \(\frac{1^2}{27^2}\)= \(\frac{1}{9}\).\(\frac{1}{729}\)= \(\frac{1}{2511}\)