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Câu 1 :
\(\frac{\left(-5\right)^{32}.20^{43}}{\left(-8\right)^{29}.125^5}\)
= \(\frac{5^{32}.2^{86}.5^{43}}{\left(-2\right)^{87}.5^{15}}\)
= \(\frac{5^{72}.\left(-2\right)^{86}}{\left(-2\right)^{87}.5^{75}}\)
= \(\frac{1}{-2}\)
Câu 2 :
\(\frac{5^4.18^4}{125.9^5.16}\)
= \(\frac{5^4.2^4.3^8}{5^3.3^{10}.2^4}\)
= \(\frac{5}{3^2}\)
= \(\frac{5}{9}\)
Câu 3 :
\(\frac{9^{18}.2^{29}}{8^9.27^{12}}\)
= \(\frac{3^{36}.2^{29}}{2^{27}.3^{36}}\)
= \(2^2\)
= 4
1.
a. \(\frac{5^4.18^4}{125.9^5.16}=\frac{5^4.2^4.3^8}{5^3.3^{10}.2^4}=\frac{5}{3^2}=\frac{5}{9}\)
2.
a . x : ( 0,25 ) 4 = ( 0,5 ) 2
=> x : ( 0,5 ) 8 = ( 0,5 ) 2
=> x = ( 0,5 ) 2 . ( 0,5 ) 8
=> x = ( 0,5 ) 10
=> x = \(\frac{1}{1024}\)
Bài 1.\(\frac{5^4\cdot18^4}{125\cdot9^5\cdot16}=\frac{5^4\cdot9^4\cdot2^4}{5^3\cdot9^5\cdot2^4}=\frac{5}{9}\)
A = ( - 5)^32 . 20^43 / ( -8 )^29 . 125^25
= 5^32 . 20^43 / ( -8 )^29 . 125^25
= 5^32 . 20^43 / ( -8 )^29 . ( 5^3 )^25
= 5^32 . 20^43 / ( -8 )^29 . 5^75
= 20^43 / ( -8 )^29 . 5^43
= ( 4 . 5 )^43 / ( -8 )^29 . 5^43
= 4^43 . 5^43 / ( -8 )^29 . 5^43
= 4^43 / ( -8 )^29
= ( 2^2 )^43 / ( -2^3 )^29
= 2^86 / -2^87
Bài 1
\(a,\left(\frac{3}{5}\right)^2-\left[\frac{1}{3}:3-\sqrt{16}.\left(\frac{1}{2}\right)^2\right]-\left(10.12-2014\right)^0\)
\(=\frac{9}{25}-\left[\frac{1}{9}-4.\frac{1}{4}\right]-1\)
\(=\frac{9}{25}-\left(-\frac{8}{9}\right)-1\)
\(=\frac{9}{25}+\frac{8}{9}-1\)
\(=\frac{56}{225}\)
\(b,|-\frac{100}{123}|:\left(\frac{3}{4}+\frac{7}{12}\right)+\frac{23}{123}:\left(\frac{9}{5}-\frac{7}{15}\right)\)
\(=\frac{100}{123}:\left(\frac{4}{3}\right)+\frac{23}{123}:\frac{4}{3}\)
\(=\left(\frac{100}{123}+\frac{23}{123}\right):\frac{4}{3}\)
\(=1:\frac{4}{3}=\frac{3}{4}\)
Phần c đăng riêng vì mk chưa tìm đc cách giải bt mỗi đáp án :v
\(c,\frac{\left(-5\right)^{32}.20^{43}}{\left(-8\right)^{29}.125^{25}}\)
\(=\frac{\left(-5\right)^{32}.\left(4.5\right)^{43}}{\left[4.\left(-2\right)\right]^{29}.\left(-5^3\right)^{25}}\)
\(=\frac{-5^{32}.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5\right)^{75}}\)
\(=\frac{\left(-5^4\right)^8.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5^3\right)^{25}}\)
\(=-\frac{1}{2}\)
a) \(A=\left(1:\frac{1}{4}\right).4+25\left(1:\frac{16}{9}:\frac{125}{64}\right):\left(-\frac{27}{8}\right)\)
\(=4.4+25.\frac{36}{125}:\frac{-27}{8}\)
\(=16-\frac{32}{15}=\frac{240}{15}-\frac{32}{15}=\frac{208}{15}\)
Câu 1:
\(\frac{5^4.18^4}{125.9^5.16}\) = \(\frac{5^4.\left(2.9\right)^4}{5^3.9^5.2^4}\) = \(\frac{5^4.2^4.9^4}{5^3.9^5.2^4}\) = \(\frac{5}{9}\)
Câu 2:
\(\frac{\left(-5\right)^{32}.20^{43}}{\left(-8\right)^{29}.125^{25}}\) = \(\frac{5^{32}.\left(4.5\right)^{43}}{\left(-2.4\right)^{29}.\left(5^3\right)^{25}}\) = \(\frac{5^{32}.4^{43}.5^{43}}{\left(-2\right)^{29}.4^{29}.5^{75}}\) = \(\frac{4^{14}.5^{43}}{\left(-2\right)^{29}.5^{43}}\)
=\(\frac{4^{14}}{\left(-2\right)^{29}}\) = = \(\frac{\left[-2.\left(-2\right)\right]^{14}}{\left(-2\right)^{29}}\) = \(\frac{\left(-2\right)^{14}.\left(-2\right)^{14}}{\left(-2\right)^{29}}\) = \(\frac{\left(-2\right)^{14}}{\left(-2\right)^{15}}\) = \(\frac{-1}{2}\)
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