về dạng luỹ thừa
.\(9.3^2.\frac{1}{81}.27\) \(4.32:\left(2^3.\frac{1}{16}\right)\)
\(3^43^5:\frac{1}{27}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(2.4.16.32.2^4=2.2^2.2^4.2^5.2^4=2^{16}\)
b) \(\left(4.2^5\right):\left(2^3.\frac{1}{16}\right)=\left(2^2.2^5\right):\left(2^3.\left(\frac{1}{2}\right)^4\right)=2^7:\frac{1}{2}=2^8\)
c) \(9.3^3.\frac{1}{81}.27=3^2.3^3.\left(\frac{1}{3}\right)^4.3^3=3^4\)
d)\(2^2.4.\frac{32}{2^2}.2^5=2^2.2^2.2^3.2^5=2^{12}\)
a/
\(9.3^2.\frac{1}{81}.27=\frac{9.3^2.27}{81}=\frac{3^2.3^2.3^3}{3^4}=\frac{3^7}{3^4}=3^3\)
b/
\(4.32:\left(2^3.\frac{1}{16}\right)=4.32:\left(\frac{2^3}{16}\right)=4.32:\left(\frac{2^3}{2^4}\right)=4.32:\frac{1}{2}=4.32.2=4.64=4.4^3=4^4\)
c/
\(3^4.3^5:\frac{1}{27}=3^4.3^5.27=3^4.3^5.3^3=3^{12}\)
d/(ý bạn là (-2)^2 hay -2^2 , mình làm theo cách (-2)^2 nhé!)
\(2^2.4.\frac{32}{\left(-2\right)^2}.2^5=2^2.2^2.\frac{2^5}{2^2}.2^5=2^2.2^2.2^3.2^5=2^{12}\)
\(a,\sqrt{2^3}=2^{\dfrac{3}{2}}\\ b,\sqrt[5]{\dfrac{1}{27}}=\sqrt[5]{3^{-3}}=3^{-\dfrac{3}{5}}\\ c,\left(\sqrt[5]{a}\right)^4=\sqrt[5]{a^4}=a^{\dfrac{4}{5}}\)
a: \(=3^2\cdot3^5:3^4=3^{2+5-4}=3^3\)
b: \(=2^3\cdot2^4:\left(\dfrac{8}{16}\right)=\dfrac{2^7}{2}=2^6\)
c: \(=3^7\cdot3^3=3^{10}\)
d: \(=5^3\cdot5^2\cdot\dfrac{1}{5^4}=5^1\)
9.32.1/81.9.3 <=> 92/81. 33= 33
4.32:(23.1/24) <=> 4.42.2.2 <=> 4.42.4 = 44
34.35:1/27 = 34.35.33= 312