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a)\(2.2.2.3.3=2^3.3^2\)
b)\(8.8.8.2=2^3.2^3.2^3.2=2^{10}\)
c)\(8^4.16^5=\left(2^3\right)^4.\left(2^4\right)^5=2^{12}.2^{20}=2^{32}\)
d)\(5^{40}.125^2.625^3=5^{40}.\left(5^3\right)^2.\left(5^4\right)^3=5^{40}.5^6.5^{12}=5^{58}\)
e)\(27^4.81^{10}=\left(3^3\right)^4.\left(3^4\right)^{10}=3^{12}.3^{40}=3^{52}\)
f)\(10^3.100^5.1000^4=10^3.\left(10^2\right)^5.\left(10^3\right)^4=10^3.10^{10}.10^{12}=10^{25}\)
a)23.22.24=2(3+2+4)=29
b)102.103.105=10(2+3+5)=1010
c)X.X5=X1.X5=X(1+5)=X6
d)a3.a2.a5=a(3+2+5)=a10
a) $3^8:3^6=3^{8-6}=3^2$
$19^7:19^3=19^{7-3}=19^4$
$2^{10}:8^3=2^{10}:(2^3)^3=2^{10}:2^9=2^{10-9}=2^1$
$12^7:6^7=(12:6)^7=2^7$
$27^5:81^3=(3^3)^5:(3^4)^3=3^{15}:3^{12}=3^{15-12}=3^3$
b) $10^6:10=10^{6-1}=10^5$
$5^8:25^2=5^8:(5^2)^2=5^8:5^4=5^{8-4}=5^4$
$4^9:64^2=4^9:(4^3)^2=4^9:4^6=4^{9-6}=4^3$
$2^25:32^4=2^{25}:(2^5)^4=2^{25}:2^{20}=2^{25-20}=2^5$
$18^3:9^3=(18:9)^3=2^3$
\(\cdot3^8:3^6=3^{8-6}=3^2\)
\(\cdot19^7:19^3=19^{7-3}=19^4\)
\(\cdot2^{10}:8^3=2^{10}:\left(2^3\right)^3=2^{10}:2^9=2\)
\(\cdot12^7:6^7=\left(12:6\right)^7=2^7\)
\(\cdot27^5:81^3=\left(3^3\right)^5:\left(3^4\right)^3=3^{15}:3^{12}=3^3\)
\(\cdot10^6:10=10^{6-1}=10^5\)
\(\cdot5^8:25^2=5^8:\left(5^2\right)^2=5^8:5^4=5^4\)
\(\cdot4^9:64^2=4^9:\left(4^3\right)^2=4^9:4^6=4^3\)
\(2^{25}:32^4=2^{25}:\left(2^5\right)^4=2^{25}:2^{20}=2^5\)
\(18^3:9^3=\left(18:9\right)^3=2^3\)
Bài 1:
a)a4.a5:a3=a4+5-3
= a6
b)m6.m.m7:m3
=m6+1+7-3
=m11
c)100.10000000:1000
=100.(107:103)
=102.107-3
=102.104
=102+4
=106
d)132-122
=13.13-12.12
=169-144
=25
=52
a. a^4 .a^5 :a^3
=\(a^9:a^3\)
=a\(^6\)
b)m6.m.m7:m3
=m\(^{14}\):m\(^3\)
=m^11
c)100.10000000:1000
=10^2.10^7:10^3
=10^9:10^3
10^6
d)13^2-12^2
=13.13-12.12
=169-144
=25
=5^2
bài 2:
a. 3(24-x^2)-10=14
3.(24-x^2)=14+10
3.(24-x^2)=24
24-x^2=24:3
24-x^2=8
x^2=24-8
x^2=16
x^2=4^2
Vậy x=4
b. (x-2)^3+2^3=6^2-1
(x-2)^3+2^3=.6.6.-1
(x-2)^3+2^3=35
(x-2)^3+8=35
(x-2)^3=35-8
(x-2)^3=27
(x-2)^3=3^3
x-2=3
x=3+2
x=5
Vậy x=5
c. 2^x+2\(^{x+3}\)=72
2^x+2\(^{x+3}\)=2^6+2^3
Vay x=6 và x=0
chúc bn học giỏi nha Thủy Trần 
a) = 53+4+1 = 58
b) = 65-4 = 6
c) = 32+3+4-7 = 32
d) = x4+2+1 = x7
e) = 812+10+3 = 825
1. \(5^3.5^4.5=5^{3+4+1}=5^8\)
2. \(6^5:\left(6^3.6\right)=6^5:\left(6^{3+1}\right)=6^5:6^4=6^{5-4}=6^1\)
3. \(\left(3^2.3^3.3^4\right):3^7=\left(3^{2+3+4}\right):3^7=3^9:3^7=3^{9-7}=3^2\)
4. \(x^4.x^2.x=x^{4+2+1}=x^7\)
5. \(\left(8^{12}.8^{10}\right).8^3=8^{12+10+3}=8^{25}\)
Bài 1 :
a) \(2^3.2^2.2^4\)
\(=2^{3+2+4}=2^9\)
b) \(10^2.10^3.10^5=10^{2+3+5}=10^{10}\)
c) \(x.x^5=x^6\)
d) \(a^3.a^2.a^5=a^{3+2+5}=a^{10}\)
Bài 2 :
a) \(c^n=>c=1\)
b) \(c^n=0=>c=0\)
Ủng hộ mik nha
Thanks
Bài 1 :
a ) \(2^3.2^2.2^4=2^{3+2+4}=2^9\)
b ) \(10^2.10^3.10^5=10^{2+3+5}=10^{10}\)
c ) \(x.x^5=x^{5+1}=x^6\)
d ) \(a^3.a^2.a^5=a^{3+2+5}=a^{10}\)
Bài 2 :A , để có c^n = 1 với n thuộc N* thì c=1. B , c^n =0 thì c = 0
a125^5:25^3=(5^3)^5:(5^2)^3=5^15:5^6=5^9
b27^6:9^3=(3^3)^6:(3^2)^3=3^18:3^6=3^13
c 4^20:2^15=(2^2)^20:2^15=2^40:2^15=2^25
d24^n:2^2.n=24^n:(2^2)^n=24^n:4^n=(24:4)^n=6^n
e 64^4 . 16^5:4^20=(2^6)^4 . (2^4)^5 :(2^2)^20=2^24 . 2^20:2^40=2^4
g 32^4:8^6=(2^5)^4:(2^3)^6=2^20:2^18=2^2
a, \(125^5:25^3=\left(5^3\right)^5:\left(5^2\right)^3=5^{15}:5^6=5^9\)
b, \(27^6:9^3=\left(3^3\right)^6:\left(3^2\right)^3=3^{18}:3^6=3^{12}\)
c, \(4^{20}:2^{15}=\left(2^2\right)^{20}:2^{15}=2^{40}:2^{15}=2^{25}\)
d, \(24^n:2^{2.n}=2^n.12^n:2^n.2^n=12^n:2^n=2^n.6^n:2^n=6^n\)
e, \(64^4.16^5:4^{20}=4^{12}.4^{10}:4^{20}=4^{12+10-20}=4^2\)
g, \(32^4:8^6=8^4.4^4:8^4.8^2=4^4:4^2.2^2=4^2.2^2=2^4.2^2=2^6\)
a) \(2^5\cdot8^4\\ =2^5\cdot\left(2^3\right)^4\\ =2^5\cdot2^{12}\\ =2^{17}\)
b) \(25^6\cdot125^3\\ =\left(5^2\right)^6\cdot\left(5^3\right)^3\\ =5^{12}\cdot5^9\\ =5^{21}\)
c) \(625^3:25^7\\ =\left(5^4\right)^3:\left(5^2\right)^7\\ =5^{12}:5^{14}\\ =5^{-2}=\frac{1}{5^2}\)
d) \(4^{10}\cdot2^{30}\\ =\left(2^2\right)^{10}\cdot2^{30}\\ =2^{20}\cdot2^{30}\\= 2^{50}\)
e) \(9^{25}\cdot27^4\cdot81^3\\ =\left(3^2\right)^{25}\cdot\left(3^3\right)^4\cdot\left(3^4\right)^3\\ =3^{50}\cdot3^{12}\cdot3^{12}\\ =3^{74}\)
a.\(2^{60}\)
b.\(10^{169}\)
c.\(x^{24}\)
d.\(10^3\)
e.\(10^9\)
f.\(10^{20}\)
a) 212
b) 1018
c) x9
d) 103
e) 109
f) 1020