viết tích sau dưới dạng một lũy thừa:
x2 . x7 . x12 . x17. x22 . ... . x492 . x497
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a, 4 8 . 4 10 = 4 18
b, 2 20 . 2 7 = 2 27
c, 5 12 . 5 5 . 5 4 = 5 21
d, 4 3 . 4 5 . 4 5 = 4 13
e, 8 6 . 8 5 . 8 5 = 8 16
f, x 7 . x 4 . x 3 = x 14
\(a,4^8.2^{20}=\left(2^2\right)^8.2^{20}=2^{16}.2^{20}=2^{16+20}=2^{36}\\ b,9^{12}.27^5.81^4=\left(3^2\right)^{12}.\left(3^3\right)^5.\left(3^4\right)^4=3^{24}.3^{15}.3^{16}=3^{24+15+16}=3^{55}\\ d,25^{20}.125^4=\left(5^2\right)^{20}.\left(5^3\right)^4=5^{40}.5^{12}=5^{40+12}=5^{52}\\ d,x^7.x^4.x^3=x^{7+4+3}=x^{14}\)
a: 4*8*2^20=2^2*2^3*2^20=2^25
b: 9^12*27^5*81^4=3^24*3^15*3^16=3^55
c: 25^20*125^4=5^40*5^12=5^52
d: =x^(7+4+3)=x^14
a: 6*6*6*6*6=6^5
b: 25*25*25*5=5^2*5^2*5^2*5=5^7
c: 8*2*2*16*4
=2^3*2^2*2^4*2^2
=2^11
d: \(3\cdot3\cdot3\cdot15\cdot5\cdot5\cdot5=15\cdot15\cdot15\cdot15=15^4\)
a: \(4^8\cdot2^{20}=\left(2^2\right)^8\cdot2^{20}=2^{16}\cdot2^{20}=2^{16+20}=2^{36}\)
\(9^{12}\cdot27^5\cdot81^4=\left(3^2\right)^{12}\cdot\left(3^3\right)^5\cdot\left(3^4\right)^4\)
\(=3^{24}\cdot3^{15}\cdot3^{16}=3^{24+15+16}=3^{55}\)
\(64^3\cdot4^5\cdot16^2=\left(4^3\right)^3\cdot4^5\cdot\left(4^2\right)^2=4^9\cdot4^5\cdot4^4=4^{9+5+4}=4^{18}\)
b: \(25^{20}\cdot125^4=\left(5^2\right)^{20}\cdot\left(5^3\right)^4=5^{40}\cdot5^{12}=5^{52}\)
\(x^7\cdot x^4\cdot x^3=x^{7+4+3}=x^{14}\)
\(3^6\cdot4^6=\left(3\cdot4\right)^6=12^6\)
c: \(8^4\cdot2^3\cdot16^2=\left(2^3\right)^4\cdot2^3\cdot\left(2^4\right)^2\)
\(=2^{12}\cdot2^3\cdot2^8=2^{23}\)
\(y\cdot y^7=y^{1+7}=y^8\)
\(2^3\cdot2^2\cdot8^3=2^5\cdot\left(2^3\right)^3=2^5\cdot2^9=2^{5+9}=2^{14}\)
a, 6 9 : 6 7 = 6 9 - 7 = 6 2
b, 7 5 : 7 2 = 7 5 - 2 = 7 3
c, 11 8 : 11 3 : 11 2 = 11 8 - 3 - 2 = 11 3
d, x 8 : x 7 : x = x 8 - 7 - 1 = x 0 = 1
x2 . x7 . x12 . x17 . x22 . ... . x492 . x497
= x2 + 7 + 12 + 17 + 22 + ... + 492 + 497
2 + 7 + 12 + 17 + 22 + ... + 492 + 497
Số số hạng : ( 497 - 2 ) : 5 + 1 = 100
Tổng : ( 497 + 2 ) . 100 : 2 = 24 950
=> x2 . x7 . x12 . x17 . x22 . ... . x492 . x497 = x24 950
Sai thì bỏ qua nhá :))