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Bài 1 :
a/ \(a^3.a^9=a^{3+9}=a^{12}\)
b/\(\left(a^5\right)^7=a^{5.7}=a^{35}\)
c/ \(\left(a^6\right).4.a^{12}=a^{24}.a^{12}.4=a^{24+12}.4=a^{36}.4\)
d/ \(\left(2^3\right)^5.\left(2^3\right)^3=2^{15}.2^9=2^{15+9}=2^{24}\)
e/ \(5^6:5^3+3^3.3^2\)
\(=5^3+3^5=125+243=368\)
i/ \(4.5^2-2.3^2\)
\(=2^2.5^2-2.3^2\)
\(=2^2.25-2^2.14\)
\(=2^2.\left(25-14\right)\)
\(=2^2.11\)
\(=4.11=44\)
Bài 1:
a) 510 : 57 - 25 . 30 = 53- 25= 125 - 32= 93
b) \(\frac{2^{14}\cdot14\cdot125}{35^3\cdot6}=\frac{2^{14}\cdot2\cdot7\cdot5^3}{\left(7\cdot5\right)^3\cdot2\cdot3}=\frac{2^{15}\cdot7\cdot5^3}{7^3\cdot5^3\cdot2\cdot3}=\frac{2^{14}}{7^2\cdot3}=\frac{16384}{49\cdot3}=\frac{16384}{174}\)
c) \(\frac{2^{13}+2^5}{2^{10}+2^2}=\frac{2^{11}\cdot2^2+2^3\cdot2^2}{2^2\cdot2^8+2^2}=\frac{2^2\left(2^{11}+2^3\right)}{2^2\left(2^8+1\right)}=\frac{2^{11}+2^3}{2^8+1}=\frac{2^3\cdot2^8+2^3}{2^8+1}=\frac{2^3\left(2^8+1\right)}{2^8+1}=2^3=8\)
Bài 2:
a) 59 . 252 = 59 . (52)2= 59 . 54= 513
b) 1410 : 495 = [(2.7)10] : [(7.7)5] = (210 . 710 ) : (75 . 75) = 210 . 710 : 710= 210
c) 414 . 528 = (22)14 . 528 = 228 . 528 = (2.5)28 = 1028
d) 10010 : (-10)3 . (-100)4 = (102)3 : 103.(-1) . (102)4= 106 : 103 . 108 .(-1) = -1011
A = 1 + 3 + 32 + 33 + ... + 3100
2A = 3 + 32 + 33 + 34 + ... + 3101
A = 2A - A = 3101 - 1
Vậy A = 3101 - 1
\(A,\frac{4^9.36+64}{16^4.100}=\frac{\left(2^2\right)^9.2^2.3^2+2^6}{\left(2^4\right)^4.2^2.5^2}=\frac{2^{20}.3^2+2^6}{2^{18}.5^2}=\frac{2^6\left(2^{14}.3^2+1\right)}{2^{18}.5^2}=\frac{2^{14}.3^2+1}{2^{12}.5^2}=\frac{147457}{102400}\)
B,
\(\frac{11.3^{22}.3-9^{13}}{\left(2.3^{14}\right)^2}=\frac{11.3^{22}-\left(3^2\right)^{13}}{2^2.3^{28}}=\frac{11.3^{22}-3^{26}}{2^2.3^{28}}=\frac{3^{22}\left(11.1-3^4\right)}{2^2.3^{28}}=\frac{11-81}{2^2.3^6}=-\frac{70}{2916}=-\frac{35}{1456}\)
c,
\(\frac{45^3.20^4.18}{180^5}=\frac{\left(3^2.5\right)^3.\left(5.2^2\right)^4.2.3^2}{\left(2^2.3^2.5\right)^5}=\frac{3^6.5^3.5^4.2^8.2.3^2}{2^{10}.3^{10}.5^5}=\frac{3^8.2^{10}.5^7}{2^{10}.3^{10}.5^5}=\frac{5^2}{3^2}=\frac{25}{9}\)
\(\frac{4^9\cdot36+64}{16^4\cdot100}=\frac{2^6\cdot147457}{2^{16}\cdot100}=\frac{147457}{2^{10}\cdot100}\)
\(\frac{11\cdot3^{22}\cdot3-9^{13}}{2^2\cdot3^{28}}=\frac{3^{23}\left(11-3^3\right)}{2^2\cdot3^{28}}=\frac{-16\cdot3^{23}}{2^2\cdot3^{28}}=\frac{-4}{243}\)
\(\frac{45^3\cdot20^4\cdot18}{180^5}=\frac{3^8\cdot2^9\cdot5^7}{2^{10}\cdot3^{10}\cdot5^5}=\frac{25}{18}\)
a)\(\left(3^2+1\right)B=\left(3^2+1\right)\cdot3\cdot\left(1-3^2+3^4-3^6+3^8-...-3^{2006}+3^{2008}\right).\)
\(10B=3\cdot\left(3^{2010}+1\right)\)
\(B=\frac{3\left(3^{2010}+1\right)}{10}\)
b) \(B=3\cdot\left(1-3^2+3^4\right)-3^7\cdot\left(1-3^2+3^4\right)+...+3^{2005}\left(1-3^2+3^4\right)\)
\(B=\left(1-3^2+3^4\right)\cdot\left(3-3^7+3^{13}-...+3^{2005}\right)=73\cdot\left(3-3^7+3^{13}-...+3^{2005}\right)\)
chia hết cho 73.
a)B=3-3^3+3^5-3^7+3^9-...+3^2009
3^2B=3^3-3^5+3^7-3^9+3^11-...+3^2011
9B+B=3^3-3^5+3^7-3^9+3^11-...+3^2011+3-3^3+3^5-3^7+3^9-...+3^2009
10B=3^2011+3
B=\(\frac{3^{2011}+3}{10}\)
b) B=3-3^3+3^5-3^7+3^9-...+3^2009
=(3-3^3+3^5)-(3^7-3^9+3^11)-....+(3^2005-3^2007+3^2009)
=(3-3^3+3^5)-[3^6(3-3^3+3^5)]-...+[3^2004(3-3^3+3^5)]
=(3-3^3+3^5)-3^6(3-3^3+3^5)-...+3^2004(3-3^3+3^5)
=219(1-3^6-...+3^2004) chia hết cho 73 vì 219 chia hết cho 73
Bài 1 :
a, ab + ba = (a*10 + b) + (b*10 + a)
= a*(10+1) + b*(1+10)
= a*11 + b*11 chia hết cho 11
b, abc - cba = (a*100 + b*10 + c) - (c*100 + b*10 + a)
= a*99 + 0b + c*(-99) chia hết cho 99
1.
4.52-2.32
=4.25-2.9
=100-18
=82