\(\frac{5^2.25^4}{125^3}=\frac{5^2.\left(5^2\right)^4}{\left(5^3\right)^3}=\frac{5^2.5^8}{5^9}=\frac{5^{10}}{5^9}=5.\)

a)\(=\frac{-7}{10}+\frac{-1}{5}=\frac{-7}{10}+\frac{-2}{10}=\frac{-9}{10}\)

b)5^x=125

5^x=5³=>x=3

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

mh biết làm bài này rùi bn có cần mi2h đang cho bn ko?

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)

\(=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)

\(=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1\)

Hay lắm bn

a)\(\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{5}{7}+\dfrac{4}{9}+\dfrac{17}{14}\)

\(=\dfrac{11}{125}-\left(\dfrac{17}{18}-\dfrac{4}{9}\right)-\left(\dfrac{5}{7}-\dfrac{17}{14}\right)\)

\(=\dfrac{11}{125}-\dfrac{5}{18}-\dfrac{-7}{14}=\dfrac{11}{125}-\dfrac{5}{18}+\dfrac{1}{2}\)

\(=\dfrac{11}{125}-\left(\dfrac{5}{18}-\dfrac{1}{2}\right)=\dfrac{11}{125}-\dfrac{-4}{18}=\dfrac{11}{125}+\dfrac{2}{9}\)