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a) \(A=\left(4n+3\right)^2-5^2=\left(4n+3-5\right)\left(4n+3+5\right)=\left(4n-2\right)\left(4n+8\right)\)
\(=8\left(n-1\right)\left(n+2\right)\). Vì A chứa thừa số 8 nên A chia hết cho 8
b) \(B=\left(2n+3\right)^2-3^2=\left(2n+3-3\right)\left(2n+3+3\right)=2n\left(2n+6\right)=4n\left(n+3\right)\)
Vì B chứa thừa số 4 nên B chia hết cho 4
a) \(A=85^2-45^2+75^2-35^2+65^2-25^2+55^2-15^2\)
\(A=\left(85-45\right)\left(85+45\right)+....+\left(55-15\right)\left(55+15\right)\)
\(A=40.130+40.110+40.90+40.70\)
\(A=40.\left(130+110+90+70\right)=40.400=16000\)
b) \(B=\left(1-2\right)\left(1+2\right)+\left(3-4\right)\left(3+4\right)+...+\left(2011-2012\right)\left(2011+2012\right)\)
\(B=-3-7-11-...-4023\)
\(B=-\left(3+7+11+...+4023\right)\)
\(B=-\dfrac{\left(3+4023\right)\left[\dfrac{\left(4023-3\right)}{4}+1\right]}{2}=2025078\)
\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{15}+1\right)\)
\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(\frac{1}{2}\left(5^{32}+1\right)=\frac{5^{32}+1}{2}\)
a)
Ta có
a chia 5 dư 4
=> a=5k+4 ( k là số tự nhiên )
\(\Rightarrow a^2=\left(5k+4\right)^2=25k^2+40k+16\)
Vì 25k^2 chia hết cho 5
40k chia hết cho 5
16 chia 5 dư 1
=> đpcm
2) Ta có
\(12=\frac{5^2-1}{2}\)
Thay vào biểu thức ta có
\(P=\frac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)}{2}\)
\(\Rightarrow P=\frac{\left[\left(5^2\right)^2-1^2\right]\left[\left(5^2\right)^2+1^2\right]\left(5^8+1\right)}{2}\)
\(\Rightarrow P=\frac{\left[\left(5^4\right)^2-1^2\right]\left[\left(5^4\right)^2+1^2\right]}{2}\)
\(\Rightarrow P=\frac{5^{16}-1}{2}\)
3)
\(\left(a+b+c\right)^3=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)
\(=a^3+b^3+c^2+3ab\left(a+b\right)+3\left(a+b\right)c\left(a+b+c\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ca+cb+c^2\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Ta có \(A=75\left(4^{1993}+4^{1992}+....+4+1\right)+25\)
\(\Leftrightarrow A=25\left(4-1\right)\left(4^{1993}+4^{1992}+...+4+1\right)+25\)
Vận dụng hằng đẳng thức
\(a^n-b^n=\left(a-b\right)\left(a^{n-1}+a^{n-2}b+...+b^{n-1}\right)\)
Ta có
\(\left(4-1\right)\left(4^{1993}+4^{1992}+...+4+1\right)=4^{1994}-1\)
\(\Rightarrow A=25\left(4^{1994}-1\right)+25\)
\(\Leftrightarrow A=25\cdot4^{1994}\)
Vậy \(A=25\cdot4^{1994}\)
đặt B = 42015 + 42014 + 42013 + ... + 42
4B = 42016 + 42015 + 42014 + ... + 43
4B - B = ( 42016 + 42015 + 42014 + ... + 43 ) - ( 42015 + 42014 + 42013 + ... + 42 )
3B = 42016 - 42
\(\Rightarrow\)B = \(\frac{4^{2016}-4^2}{3}\)hay B = \(\frac{4^{2016}-16}{3}\)
\(\Rightarrow\)A = 75 . ( \(\frac{4^{2016}-16}{3}\)+ 5 ) + 25
A = 75 . ( \(\frac{4^{2016}-16}{3}\)+ \(\frac{15}{3}\)) + 25
A = 75 . ( \(\frac{4^{2016}-1}{3}\)) + 25
A = 25 . ( 3 . \(\frac{4^{2016}-1}{3}\)) + 25
A = 25 . ( 42016 - 1 ) + 25
A = 25 . ( 42016 - 1 + 1 )
A = 25 . 42016 \(⋮\)42016
\(A=25.3\left(4^{1975}+4^{1974}+...+4^2+4+1\right)+25\)
\(=25\left(4-1\right)\left(4^{1975}+4^{1974}+...+4^2+4+1\right)+25\)
Áp dụng hằng đẳng thức, ta có : \(A=25\left(4^{1976}-1\right)+25=25.4^{1976}\)
Vậy \(A⋮4^{1976}\)
banj áp dụng hằng đẳng thức nào vậy ạ??