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

mk k bít hihi

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

AI GIÚP MÒNH VỚI !

k rồi giúp

A=3/4

B=29/35

mình nghĩ thế

Sai thôi nhé!

\(A=\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

\(A=\left(\frac{3}{8}+\frac{-6}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

\(A=\left(\frac{-3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

\(A=\left(\frac{-36}{24}+\frac{56}{24}\right):\frac{5}{6}+\frac{1}{2}\)

\(A=\frac{5}{6}:\frac{5}{6}+\frac{1}{2}\)

\(A=\frac{5}{6}\times\frac{6}{5}+\frac{1}{2}\)

\(A=1+\frac{1}{2}\)

\(A=\frac{1}{1}+\frac{1}{2}=\frac{2}{2}+\frac{1}{2}\)

\(A=\frac{3}{2}\)

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 

\(c,\frac{\left(-5\right)^{32}.20^{43}}{\left(-8\right)^{29}.125^{25}}\)

\(=\frac{\left(-5\right)^{32}.\left(4.5\right)^{43}}{\left[4.\left(-2\right)\right]^{29}.\left(-5^3\right)^{25}}\)

\(=\frac{-5^{32}.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5\right)^{75}}\)

\(=\frac{\left(-5^4\right)^8.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5^3\right)^{25}}\)

\(=-\frac{1}{2}\)

Bài 2

\(a,\frac{2}{3}-\left(\frac{3}{4}-x\right)=\sqrt{\frac{1}{9}}\)

\(\frac{2}{3}-\left(\frac{3}{4}-x\right)=\frac{1}{3}\)

\(\frac{3}{4}-x=\frac{2}{3}-\frac{1}{3}\)

\(\frac{3}{4}-x=\frac{1}{3}\)

\(x=\frac{3}{4}-\frac{1}{3}\)

\(x=\frac{5}{12}\)

\(b,\left(\frac{1}{2}-x\right)^2=\left(-2\right)^2\)

\(\Rightarrow\frac{1}{2}-x=-2\) hoặc                   \(\frac{1}{2}-x=2\)

\(\Rightarrow x=\frac{1}{2}-\left(-2\right)\)                      \(x=\frac{1}{2}-2\)

\(\Rightarrow x=\frac{5}{2}\)                                         \(x=-\frac{3}{2}\)

\(c,|3x+\frac{1}{2}|-\frac{2}{3}=1\)

\(|3x+\frac{1}{2}|=1+\frac{2}{3}\)

\(|3x+\frac{1}{2}|=\frac{5}{3}\)

\(\Rightarrow\orbr{\begin{cases}3x+\frac{1}{2}=\frac{5}{3}\\3x+\frac{1}{2}=-\frac{5}{3}\end{cases}}\Rightarrow\orbr{\begin{cases}3x=\frac{5}{3}-\frac{1}{2}=\frac{7}{6}\\3x=-\frac{5}{3}-\frac{1}{2}=-\frac{13}{6}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{18}\\x=-\frac{13}{18}\end{cases}}}\)

\(d,3^{2x+1}=81\)

\(\Rightarrow3^{2x+1}=3^4\)

\(\Rightarrow2x+1=4\)

\(\Rightarrow2x=3\)

\(\Rightarrow x=\frac{3}{2}\)