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a, \(3^{15}:3^5=3^{15-5}=3^{10}\)
b, \(4^6:4^6=4^{6-6}=4^0\)
c, \(9^8:3^2=9^8:9=9^{8-1}=9^7\)
a) \(3^8:3^4=3^{8-4}=3^4\)
b) \(10^8:10^2=10^{8-2}=10^6\)
c) \(a^6:a=a^{6-1}=a^5\)
Áp dụng quy tắc am : an = am - n(a ≠ 0, m ≥ n ).
a) 38 : 34 = 38 – 4 = 34 = 81;
b) 108 : 102 = 108 – 2 = 106 = 1000000
c) a6 : a = a6 – 1 = a5
a) \(3^{7} \cdot 27^{5} \cdot 81^{3}\)
\(3^{7} \cdot \left(\right. 3^{3} \left.\right)^{5} \cdot \left(\right. 3^{4} \left.\right)^{3} = 3^{7} \cdot 3^{15} \cdot 3^{12} = 3^{7 + 15 + 12} = 3^{34} .\)
b) \(100^{6} \cdot 1000^{5} \cdot 10 \textrm{ } 000^{3}\)
\(\left(\right. 10^{2} \left.\right)^{6} \cdot \left(\right. 10^{3} \left.\right)^{5} \cdot \left(\right. 10^{4} \left.\right)^{3} = 10^{12} \cdot 10^{15} \cdot 10^{12} = 10^{39} .\)
c) \(\frac{36^{5}}{18^{5}}\)
\(\left(\left(\right. \frac{36}{18} \left.\right)\right)^{5} = 2^{5} = 32.\)
d) \(24 \cdot 5^{2} + 5^{2} \cdot 5^{3}\)
\(= 24 \cdot 25 + 25 \cdot 125 = 600 + 3125 = 3725.\)
e) \(\frac{125^{4}}{5^{8}}\)
\(\left(\right. 5^{3} \left.\right)^{4} : 5^{8} = 5^{12} : 5^{8} = 5^{12 - 8} = 5^{4} = 625.\)
TÓM lại là a) \(3^{34}\)
b) \(10^{39}\)
c) \(32\)
d) \(3725\)
e) \(625\)
bạn muốn chép đáp án hay sem cách làm///??
a: \(3^7\cdot27^5\cdot81^3\)
\(=3^7\cdot\left(3^3\right)^5\cdot\left(3^4\right)^3\)
\(=3^7\cdot3^{15}\cdot3^{12}=3^{7+15+12}=3^{34}\)
b: \(100^6\cdot1000^5\cdot10000^3\)
\(=\left(10^2\right)^6\cdot\left(10^3\right)^5\cdot\left(10^4\right)^3\)
\(=10^{12}\cdot10^{15}\cdot10^{12}=10^{15+12+12}=10^{39}\)
c: \(36^5:18^5=\left(\frac{36}{18}\right)^5=2^5=32\)
d: \(24\cdot5^2+5^2\cdot5^3\)
\(=5^2\left(24+5^3\right)\)
\(=25\cdot\left(24+125\right)=25\cdot149=3725\)
e: \(125^4:5^8=\left(5^3\right)^4:5^8=5^{12}:5^8=5^{12-8}=5^4\)
a) a3.a5 = a3+5 = a8
b) x7.x.x4 = x7+1+4 = x12
c) 35.45 = (3.4)5 = 125
d) 85.23 = 25.25.25.23 = 25+5+5+3 = 218
a/a\(^3\).a\(^5\)=a\(^8\)
b/x\(^7\).x.x\(^4\)=x\(^{12}\)
c/3\(^5\).4\(^5\)=2\(^{10}\).3\(^5\)
d/8\(^5\).2\(^3\)=2\(^{18}\)
99,
\(a,3^2+4^2=9+16=25\)
Mà 25 là số chính phương nên \(3^2+4^2\)là số chính phương
\(b,5^2+12^2=25+144=169\)
Mà 169 là số chính phương nên \(5^2+12^2\)là số chính phương
100,
a, \(3^{15}:3^5=3^{15-5}=3^{10}\)
b\(4^6:4^6=1\)
c\(9^8:3^2=9^{4.2}:3^2=6561^2:3^2=\left(6561-3\right)^2=6558^2\)
Theo quy tắc nhân hai lũy thừa cùng cơ số: am.an = am + n ta có:
a) 33.34 = 37 b) 52.57 = 59; c) 75.7 = 76.
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 1b)
\(8\frac{1}{14}-6\frac37\)
C1:\(\frac{113}{14}-\frac{45}{7}\) =\(\frac{113}{14}-\frac{90}{14}=\frac{23}{14}\)
C2:\(8\frac{1}{14}-6\frac37=\left(8-6\right)+\left(\frac{1}{14}-\frac37\right)=2+\left(\frac{1}{14}-\frac{6}{14}\right)\)
\(=2+\frac{-5}{14}=\frac{28}{14}-\frac{5}{14}=\frac{23}{14}\)
bài 1 c)\(7-3\frac67\)
C1:\(\) \(7-3\frac67=7-\frac{27}{7}=\frac{49}{7}-\frac{27}{7}=\frac{22}{7}\)
C2:\(7-3\frac67=\left(7-3\right)-\frac67=4-\frac67=\frac{28}{7}-\frac67=\frac{22}{7}\)
