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a, \(125^3:5^7=\left(5^3\right)^3:5^7=5^9:5^7=5^2\)
b, \(\left(\dfrac{2}{7}\right)^{18}:\left(\dfrac{4}{49}\right)^5:\left(\dfrac{8}{343}\right)^2\)
= \(\left(\dfrac{2}{7}\right)^{18}:\left(\dfrac{2^2}{7^2}\right)^5:\left(\dfrac{2^3}{7^3}\right)^2\)
= \(\left(\dfrac{2}{7}\right)^{18}:\left[\left(\dfrac{2}{7}\right)^2\right]^5:\left[\left(\dfrac{2}{7}\right)^3\right]^2\)
=\(\left(\dfrac{2}{7}\right)^{18}:\left(\dfrac{2}{7}\right)^{10}:\left(\dfrac{2}{7}\right)^6\)
= \(\left(\dfrac{2}{7}\right)^{18-10-6}=\left(\dfrac{2}{7}\right)^2\)
c, \(3-\left(\dfrac{-7}{9}\right)^0+\left(\dfrac{1}{3}\right)^5.3^5\)
= 3 - 1 +\(\left[\left(\dfrac{1}{3}\right)^5.3^5\right]\)
= 2 + 1=3
d, \(\dfrac{45^{10}.5^{20}}{75^{15}}=\dfrac{\left(9.5\right)^{10}.5^{20}}{\left(25.3\right)^{15}}=\dfrac{\left(3^2\right)^{10}.5^{10}.5^{20}}{\left(5^2\right)^{15}.3^{15}}\)
= \(\dfrac{3^{20}.5^{30}}{5^{30}.3^{15}}=3^5\)
a: \(=\left(-\dfrac{5}{7}\right)^{n-n}=\left(-\dfrac{5}{7}\right)^0=1\)
b: \(=\left(-\dfrac{1}{2}\right)^{2n-n}=\left(-\dfrac{1}{2}\right)^n\)
a) \(\dfrac{\left(-\dfrac{5}{7}\right)^n}{\left(-\dfrac{5}{7}\right)^{n-1}}\)
\(=\dfrac{\left(-\dfrac{5}{7}\right)^n}{\left(-\dfrac{5}{7}\right)^n:\left(-\dfrac{5}{7}\right)}\)
\(=\dfrac{\left(-\dfrac{5}{7}\right)^n}{\left(-\dfrac{5}{7}\right)^n.\left(-\dfrac{7}{5}\right)}\)
\(=\dfrac{1}{\left(-\dfrac{7}{5}\right)}\)
\(=1.\left(-\dfrac{5}{7}\right)\)
\(=-\dfrac{5}{7}\)
b) \(\dfrac{\left(-\dfrac{1}{2}\right)^{2n}}{\left(-\dfrac{1}{2}\right)^n}\)
\(=\dfrac{\left(-\dfrac{1}{2}\right)^n.\left(-\dfrac{1}{2}\right)^n}{\left(-\dfrac{1}{2}\right)^n}\)
\(=\left(-\dfrac{1}{2}\right)^n\)
1. Tìm n, biết:
a) \(\dfrac{-32}{\left(-2\right)^n}=4\)
\(\Rightarrow\dfrac{\left(-2\right)^5}{\left(-2\right)^n}=\left(-2\right)^2\)
\(\Rightarrow\left(-2\right)^n.\left(-2\right)^2=\left(-2\right)^5\)
(-2)n + 2 = (-2)5
n + 2 = 5
n = 5 - 2
n = 3.
b) \(\dfrac{8}{2^n}=2\)
\(\Rightarrow\dfrac{2^3}{2^n}=2\)
\(\Rightarrow\) 2n . 2 = 23
n + 1 = 3
n = 3 - 1
n = 2.
c) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3\)
2n - 1 = 3
2n = 3 + 1
2n = 4
n = 4 : 2
n = 2.
2. Tính:
a) \(\left(\dfrac{1}{2}\right)^3.\left(\dfrac{1}{4}\right)^2\)
\(=\left(\dfrac{1}{2}\right)^3.\left[\left(\dfrac{1}{2}\right)^2\right]^2\)
\(=\left(\dfrac{1}{2}\right)^3.\left(\dfrac{1}{2}\right)^4\)
\(=\left(\dfrac{1}{2}\right)^7\)
\(=\dfrac{1}{128}\)
b) 273 : 93
= (33)3 : (32)3
= 39 : 36
= 33
= 27
c) 1252 : 253
= (53)2 : (52)3
= 56 : 56
= 1
d) \(\dfrac{27^2.8^5}{6^6.32^3}\)
\(=\dfrac{\left(3^3\right)^2.\left(2^3\right)^5}{6^6.\left(2^5\right)^3}\)
\(=\dfrac{3^6.2^{15}}{6^6.2^{15}}\)
\(=\dfrac{3^6}{6^6}\)
\(=\dfrac{1}{64}.\)
B2 :
b) 27\(^3\): 9\(^3\)= (27:9)\(^3\)= 3\(^3\)
c) 125\(^2\): 25\(^3\)= 15625 : 15625 = 1
a)\(\left(\dfrac{1}{2}\right)^n=\dfrac{1}{32}\)
=>\(\left(\dfrac{1}{2}\right)^n=\left(\dfrac{1}{2}\right)^5\)
=>n=5
b)\(\left(\dfrac{343}{125}\right)=\left(\dfrac{7}{5}\right)^n\)
=>\(\left(\dfrac{7}{5}\right)^3=\left(\dfrac{7}{5}\right)^n\)
=>n=3
c)\(\dfrac{16}{2^n}=2\)
=>2n=\(\dfrac{16}{2}\)
=>2n=8
=>2n=23
=>n=3
d)\(\dfrac{\left(-3\right)^n}{81}=-27\)
=>(-3)n=-27.81
=>(-3)n=-2187
=>(-3)n=(-3)7
=>n=7
e)8n:2n=4
=>(23)n:2n=4
=>23n:2n=4
=>23n-n=4
=>22n=4
=>22n=22
=>2n=2
=>n=1
f)32.3n=35
=>3n=35:32
=>3n=35-2
=>3n=33
=>n=3
g) (22:4).2n=4
=>1.2n=22
=>n=2
h)3-2.34.3n=37
=>\(\left(\dfrac{1}{3}\right)^2\).34.3n=37
=>32.3n=37
=>32+n=37
=>2+n=7
=>n=5
a) \(2^{-1}\cdot2^n+4\cdot2^n=9\cdot2^5\)
\(\Rightarrow2^n\cdot\left(2^{-1}+4\right)=9\cdot2^5\)
\(\Rightarrow2^n\cdot4,5=288\)
\(\Rightarrow2^n=64\)
\(\Rightarrow n=6\)
b) \(2^m-2^n=1984\)
\(\Rightarrow2^n\cdot\left(2^{m-n}-1\right)=2^6\cdot31\)
\(\Rightarrow\left\{{}\begin{matrix}2^n=2^6\\2^{m-n}-1=31\end{matrix}\right.\)
\(\Rightarrow n=6\)
\(\Rightarrow2^{m-n}=32\Rightarrow m-n=5\Rightarrow m=11\)
a) (12)m=132(12)m=132
\(\Rightarrow\left(\dfrac{1}{2}\right)^m=\left(\dfrac{1}{2}\right)^5\Rightarrow m=5\)
b)
343125=(75)n
\(\Rightarrow\left(\dfrac{7}{5}\right)^3=\left(\dfrac{7}{5}\right)^n\Rightarrow n=3\)
a) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3\)
\(\Rightarrow2n-1=3\)
\(\Rightarrow2n=4\)
\(\Rightarrow n=2\)
a) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)
\(\Rightarrow2^{-\left(2n-1\right)}=2^{-3}\)
\(\Rightarrow2^{-2n+1}=2^{-3}\)
\(\Rightarrow-2n+1=-3\)
\(\Rightarrow-2n=-4\)
\(\Rightarrow n=-2\)
Vậy ...
b) \(\left(\dfrac{7}{5}\right)^n=\dfrac{343}{125}\)
\(\Rightarrow\left(\dfrac{7}{5}\right)^n=\left(\dfrac{7}{5}\right)^3\)
\(\Rightarrow n=3\)
Vậy ....