5^40 . 125 ^ 7
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\(\frac{1}{5}+\frac{1}{7}+\frac{1}{3}+\frac{100}{125}+\frac{66}{77}+\frac{40}{60}\)
\(=\frac{1}{5}+\frac{1}{7}+\frac{1}{3}+\frac{4}{5}+\frac{6}{7}+\frac{2}{3}\)
\(=\left(\frac{1}{5}+\frac{4}{5}\right)+\left(\frac{1}{7}+\frac{6}{7}\right)+\left(\frac{1}{3}+\frac{2}{3}\right)\)
\(=1+1+1\)
\(=3\)
câu 1 : ( 26 - 3x ) : 5 + 71 = 75
( 26 - 3x ) : 5 = 4
26 - 3x =20
3x = 26 - 20
3x=6
x=6: 3
x=2
1/2×(2×x-3)+105/2=-137/2
1/2×(2×x-3)=-137/2-105/2
1/2×(2×x-3)=-242/2
2×x-3=-242/2:1/2
2×x-3=-242
2.x=(-242)+3
2.x=239
x=239:2
x=239/2
A,-12×25-25×75-25×13
=[(-12)-75-13].25
=(-100).25
-2500
B,-50/7×49/10-35/2×-10/7+ -25/3× -9/5
=(-35)-(-25)+(-13)
=-23
C,-354+265-156+125
=[(-354)-156]+(265+125)
=(-510)+277
=-233
D,-74+40-50+16-35
=(-74)+40+(-50)+16+(-35)
=[(-74)+(-35)]+[40+(-50)+16]
=(-109)+26
=-83
E,-2/3×4/5+-4/5×4/3-0,125
=-2/3.4/5+4/5.(-4/3)-0,125
=4/5.[-2/3+(-4/3)]-0,125
=4/5.(-2)-0,125
=-8/5-0,125
=(-1,6)+(-0,125)
=-1,725
a) \(\left(\dfrac{1}{2}\right)^n=\dfrac{1}{32}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^n=\dfrac{1^5}{2^5}\)
\(\Rightarrow n=5\)
Vậy n = 5
c) \(\dfrac{1}{9}\cdot27=3^n\)
\(\Rightarrow3=3^n\)
\(\Rightarrow n=1\)
Vậy n = 1
a) \(125^5=\left(5^3\right)^5=5^{3\cdot5}=5^{15}\)
\(25^7=\left(5^2\right)^7=5^{2\cdot7}=5^{14}\)
\(5^{15}>5^{14}\Rightarrow125^5>25^7\)
b) \(3^{54}=\left(3^2\right)^{27}\)
\(2^{81}=\left(2^3\right)^{27}\)
\(3^2>2^3\Rightarrow3^{54}>2^{81}\)
d)
5^40 = ( 5^4)^10 = 625^10
mà 625^10 > 620^10 => 5^40 > 620^10
vậy ............
c)
10^30 = (10^3)^10 = 1000^10
2^100 = (2^10)^10 = 1024^10
mà 1000^10 < 1024^10 => 10^30 < 2^100
k mik nha!
a: \(16=2^4\)
nên \(-\dfrac{5}{16}\) viết được dưới dạng số thập phân hữu hạn
\(-\dfrac{5}{16}=-0.3125\)
\(5^{40}\cdot125^7\)
\(=5^{40}\cdot\left(5^3\right)^7\)
\(=5^{40}\cdot5^{21}\)
\(=5^{40+21}\)
\(=5^{61}\)
5 ^ 40 . 125^7
= 5 ^ 40 . ( 5^3)^7
= 5^40 . 5^3.7
= 5^40 . 5^21
= 5^61