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1.
a) x : \(\left(\dfrac{3}{4}\right)^3\) =\(\left(\dfrac{3}{4}\right)^3\)
x = \(\left(\dfrac{3}{4}\right)^3.\left(\dfrac{3}{4}\right)^3\)
x = \(\dfrac{3}{4}^{3+3}\)
x = \(\dfrac{3}{4}^6\)
x = \(\dfrac{729}{4096}\)
b) \(\left(\dfrac{2}{5}\right)^5.x=\left(\dfrac{2}{5}\right)^8\)
x = \(\left(\dfrac{2}{5}\right)^8:\left(\dfrac{2}{5}\right)^5\)
x = \(\dfrac{2}{5}^{8-5}\)
x = \(\dfrac{2}{5}^3\)
x = \(\dfrac{8}{5}\)
2.
(0,36)\(^8\) \([\left(0,6\right)^3]^8\) = (0,6)\(^{3.8}\) = ( 0,6)\(^{24}\)
( 0,216)\(^4\) = \([\left(0,6\right)^3]^4\) = (0.6)\(^{3.4}\) = ( 0,6)\(^{12}\)
\(x:\left(\dfrac{3}{4}\right)^3=\left(\dfrac{3}{4}\right)^2\)
\(x=\left(\dfrac{3}{4}\right)^2.\left(\dfrac{3}{4}\right)^3\) <=> \(x=\left(\dfrac{3}{4}\right)^{2+3}\)
=> \(x=\left(\dfrac{3}{4}\right)^5\)
b, \(\left(\dfrac{2}{5}\right)^5.x=\left(\dfrac{2}{5}\right)^8\)
\(x=\left(\dfrac{2}{5}\right)^8:\left(\dfrac{2}{5}\right)^5\Leftrightarrow x=\left(\dfrac{2}{5}\right)^{8-5}\)
=>\(x=\left(\dfrac{2}{5}\right)^3\)
bài 2 : Với bài này ta cần áp dụng quy tắc: \(\left(x^m\right)^n=x^{m.n}\)
\(0,36^8=\left[\left(0,6\right)^2\right]^8=\left(0,6\right)^{16}\)
\(0,216^4=\left[\left(0,6\right)^3\right]^4=\left(0,6\right)^{12}\)
a: \(=2^2\cdot9\cdot\dfrac{1}{6\cdot9}\cdot\dfrac{4^2}{9^2}=\dfrac{2^2}{6}\cdot\dfrac{2^4}{3^4}=\dfrac{2^6}{2\cdot3\cdot3^4}=\dfrac{2^5}{3^5}=\left(\dfrac{2}{3}\right)^5\)
b: \(=2^8\cdot\dfrac{3^4}{2^4}=3^4\cdot2^4=6^4\)
c: \(=\dfrac{\left(\dfrac{1}{2}\right)^3\cdot2^3\cdot\left(\dfrac{1}{2}\right)^2}{\left(-8\right)^2\cdot16}\cdot2^6=\dfrac{\dfrac{1}{2^2}}{64\cdot16}\cdot64=\dfrac{1}{4}:16=\dfrac{1}{64}=\left(\dfrac{1}{8}\right)^2\)
a: \(=2^2\cdot9\cdot\dfrac{1}{3^3\cdot2}\cdot\dfrac{2^4}{3^4}=\dfrac{2^4\cdot2^2}{2}\cdot\dfrac{9}{3^3\cdot3^4}=\dfrac{2^5}{3^5}=\left(\dfrac{2}{3}\right)^5\)
b: \(=2^8\cdot\dfrac{3^4}{2^4}=3^4\cdot2^4=6^4\)
c: \(=\dfrac{\dfrac{1}{2^3}\cdot\dfrac{1}{2^2}\cdot8}{\left(-8\right)^2\cdot2^4}\cdot2^6=\dfrac{1}{2^2}\cdot2^6:2^{10}=\dfrac{2^4}{2^{10}}=\dfrac{1}{2^6}=\left(\dfrac{1}{8}\right)^2\)
\(\dfrac{81}{125}=\dfrac{3^4}{5^3}\) ; \(-\dfrac{8}{27}=-\left(\dfrac{2}{3}\right)^3\)
Bài 1:
a) \(3^7:3^5-\left(\dfrac{5}{17}\right)^0=3^{7-5}-1=3^2-1=9-1=8\)
b) \(\left(\dfrac{5}{2}\right)^{13}:\left(\dfrac{1}{2}+2\right)^3\)
\(=\left(\dfrac{5}{2}\right)^{13}:\left(\dfrac{5}{2}\right)^3\)
\(=\left(\dfrac{5}{2}\right)^{10}\)
c) \(8.\left(\dfrac{1}{4}\right)^3+\left(\dfrac{2}{27}\right)^0-\dfrac{1}{8}\)
\(=8.\dfrac{1}{64}+1-\dfrac{1}{8}\)
\(=\dfrac{1}{8}+1-\dfrac{1}{8}\)
\(=1\)
Bài 2:
a) \(\dfrac{3^4.4^4}{6^4}=\dfrac{3^4.\left(2^2\right)^4}{\left(2.3\right)^4}=\dfrac{3^4.2^8}{2^4.3^4}=\dfrac{2^8}{2^4}=2^4=16\)
b) \(\dfrac{15^3}{10^3}=\dfrac{\left(3.5\right)^3}{ \left(2.5\right)^3}=\dfrac{3^3.5^3}{2^3.5^3}=3^3:2^3=\dfrac{27}{8}\)
c) \(\dfrac{4^2.12^5}{9^2.2^{10}}=\dfrac{\left(2^2\right)^2.\left[3.\left(2^2\right)\right]^5}{\left(3^2\right)^2.2^{10}}=\dfrac{2^4.3^5.2^{10}}{3^4.2^{10}}=2^4.3=16.3=48\)
d) \(\dfrac{6^2+5.2^2+4}{15}=\dfrac{\left(2.3\right)^2+5.2^2+2^2}{15}=\dfrac{2^2.3^2+5.2^2+2^2}{15}=\dfrac{2^2\left(3^2+5+1\right)}{15}=\dfrac{2^2.15}{15}=2^2=4\)
Bài 3:
a) \(\dfrac{\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.\left(-1\right)^5}{\left(\dfrac{2}{5}\right)^2.\left(\dfrac{-5}{12}\right)^2}\)
\(=\dfrac{\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.-1}{\left[\dfrac{2}{5}.\left(\dfrac{-5}{12}\right)\right]^2}\)
\(=\dfrac{\left(\dfrac{2}{3}\right)^3. \left(\dfrac{-3}{4}\right)^2.-1}{\left(\dfrac{-1}{6}\right)^2}\)
\(=\left(\dfrac{2}{3}\right)^3.\left[\left(\dfrac{-3}{4}\right).-6\right]^2.-1\)
\(=\left(\dfrac{2}{3}\right)^3.\left(\dfrac{9}{2}\right)^2.-1\)
\(=\left(\dfrac{2}{3}\right)^2.\dfrac{2}{3}.\left(\dfrac{9}{2}\right)^2.-1\)
\(=\left(\dfrac{2}{3}.\dfrac{9}{2}\right)^2.\dfrac{2}{3}.-1\)
\(=9.\dfrac{2}{3}.-1\)
\(=6.-1=-6\)
b) \(\dfrac{6^6+6^3.3^3+3^6}{-73}=\dfrac{\left(2.3\right)^6+\left(2.3\right)^3.3^3+3^6}{-73}=\dfrac{2^6.3^6+2^3.3^3.3^3+3^6}{-73}=\dfrac{2^6.3^6+2^3.3^6+3^6}{-73}=\dfrac{3^6\left(2^6+2^3+1\right)}{-73}=\dfrac{3^6.73}{-73}=\dfrac{3^6}{-1}=\left(-3\right)^6\)
\(#Wendy.Dang\)
Lần sau bnn gửi từng bài thôi nha, chứ như vầy nhiều quá thì làm không nổi mất. đánh máy nãy giờ lú luôn gòi nè :))
Bài 2: \(\dfrac{81}{225}=\dfrac{9^2}{15^2}=\left(\dfrac{9}{15}\right)^2\)
\(\left(-\dfrac{8}{27}\right)=\left(-\dfrac{2^3}{3^3}\right)=\left(-\dfrac{2}{3}\right)^3\)
Tgame96 :)))