Các bn giúp mk vs nhé trc tối mai đc k thanks trc nha
1, Viết các biểu thức dưới dạng một lũy thừa:
a,35.. 3. 310 : 315
b, 48. 25. 83
c, 162 : 43
d, 2254: 58
e, 3616 :( - 18)10
2,Tìm x,biết:
a,x2- 22 = 32
b,x3+ 5 =4
c, x3- 4.x= 0
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a, 48.84
= (22)8.(23)4
= 216.212
= 228
b, 415.515
= (4.5)15
= 2015
c, 210.15 + 210.85
= 210.(15 + 85)
= 210.100
=210.(2.5)2
= 212.52
d, 33.92
= 33 . (32)2
= 33.34
= 37
e, 512.7 - 511.10
= 511.(5.7 - 10)
= 511.25
=511.52
=513
f, \(x^1\).\(x^2\).\(x^3\)....\(x^{100}\)
= \(x^{1+2+3+...+100}\)
= \(x^{\left(1+100\right).100:2}\)
= \(x^{5050}\)
(2a3x2y).(8a2x3y4).(16a3x3y3)
= 2a3 . x3 . 16a3 . x3 .y3 . x2.8a2 .y4
= a3x3a3 . x3 .y3 .2.16.8. x2.a2y4
=(axaxy)3 .y4 44 .(xy)2
=(axaxy)3 .(4y)4 (xy)2
a: \(2^6\cdot3^3=\left(2^2\cdot3\right)^3=12^3\)
b: \(6^4\cdot8^3=2^4\cdot3^4\cdot2^9=2^{13}\cdot3^4\)
c: \(16\cdot81=36^2\)
d: \(25^4\cdot2^8=100^4\)
Bài 1:
a) \(4^8\cdot2^{20}=\left(2^2\right)^8\cdot2^{20}=2^{36}\)
\(64^3\cdot4^5=\left(2^6\right)^3\cdot\left(2^2\right)^5=2^{18}\cdot2^{10}=2^{28}\)
\(y\cdot y^7=y^{1+7}=y^8\)
\(a^n\cdot a^2=a^{n+2}\)
Bài 1:
b) \(10^8:2^8=5^8\)
\(17^8:17^5=17^3\)
\(2^{25}:32^4=2^{25}:2^{20}=2^5\)
\(19^4:9^4=\left(\dfrac{19}{9}\right)^4\)
\(a,10^8.2^8=\left(10.2\right)^8=20^8\)
\(b,10^8:2^8=\left(10:2\right)^8=5^8\)
\(c,25^4.2^8=25^4.\left(2^2\right)^4=25^4.4^4=\left(25.4\right)^4=100^4\)
\(d,15^8.9^4=15^8.\left(3^2\right)^4=15^8.3^8=\left(15.3\right)^8=45^8\)
\(e,27^2.25^3=\left(3^3\right)^2.\left(5^2\right)^3=3^6.5^6=\left(3.5\right)^6=15^6\)
a, 108 . 28
= ( 10 . 2 ) 8
= 208
b, 108 : 28
= ( 10 : 2 )8
= 58
c, 254 . 28
= ( 52 )4 . 28
= 58 . 28
= ( 5 . 2 ) 8
= 108
d, 158 . 94
= 158 . ( 32 )4
= 158 . 38
= ( 15 . 3 ) 8
= 458
e, 272 . 253
= ( 33 )2 . ( 52 ) 3
= 36 . 56
= ( 3 . 5 ) 6
= 15 6
\(a)3^5.3.3^{10}:3^{15}=3^{5+1+10-15}=3\)
\(b)4^8.2^5.8^3=\left(2^2\right)^8.2^5.\left(2^3\right)^3=2^{16}.2^5.2^9=2^{16+5+9}=2^{30}\)
\(c)16^2:4^3=\left(4^2\right)^2:4^3=4^4:4^3=4\)
a,x2- 22 = 32
⇔ x2=32+22
⇔ x2=36
⇔ x= \(\pm6\)
vậy x=\(\pm6\)
b,x3+ 5 =4
⇔ x3=4-5
⇔ x3=-1
⇔ x=-1
vậy x=-1
c, x3- 4.x= 0
⇔ x(x2-4)=0
⇔ x(x-2)(x+2)=0
⇔ \(\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
vậy .....