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a) \(-6.\left(-\frac{2}{3}\right).0,25\)
\(=-6.\left(-\frac{2}{3}\right).\frac{1}{4}\)
\(=4.\frac{1}{4}\)
\(=1.\)
b) \(-\frac{15}{4}.\left(-\frac{7}{15}\right).\left(-2\frac{2}{5}\right)\)
\(=-\frac{15}{4}.\left(-\frac{7}{15}\right).\left(-\frac{12}{5}\right)\)
\(=\frac{7}{4}.\left(-\frac{12}{5}\right)\)
\(=-\frac{21}{5}.\)
c) \(\left(-0,4\right)^2-\left(-0,4\right)^3.\left(-3\right)\)
\(=0,16-\left(-0,064\right).\left(-3\right)\)
\(=0,16-0,192\)
\(=-0,032.\)
Chúc bạn học tốt!

a.
\(\frac{45^{10}\times5^{20}}{75^{15}}=\frac{\left(3^2\times5\right)^{10}\times5^{20}}{\left(3\times5^2\right)^{15}}=\frac{3^{20}\times5^{10}\times5^{20}}{3^{15}\times5^{30}}=3^5=243\)
b.
\(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(0,8\right)^5}{\left(0,4\right)^5}\times\frac{1}{\left(0,4\right)}=\left(\frac{0,8}{0,4}\right)^5\times\frac{1}{\frac{4}{10}}=2^5\times\frac{5}{2}=2^4\times5=16\times5=80\)
c.
\(\frac{2^{15}\times9^4}{6^6\times8^3}=\frac{2^{15}\times\left(3^2\right)^4}{\left(2\times3\right)^6\times\left(2^3\right)^3}=\frac{2^{15}\times3^8}{2^6\times3^6\times2^9}=3^2=9\)
Chúc bạn học tốt ^^

\(\dfrac{\text{45^{10^{ }}}.5^{10}}{75^{10}}=\dfrac{9^{10}.5^{10}.5^{10}}{5^{10}.5^{10}.3^{10}}=\dfrac{9^{10}}{3^{10}}=3^{10}\)
\(\dfrac{\left(0,8\right)^5}{\left(0,4\right)^6}=\dfrac{2^5.\left(0,4\right)^5}{\left(0,4\right)^6}=\dfrac{2^5}{0,4}=\dfrac{32}{0,4}=80\)

a, \(3|x-0,5|-2x=x+0,4.\)
\(\Leftrightarrow3|x-0,5|=3x+0,4\)
\(\Leftrightarrow|x-0,5|=x+0,4\)
\(\Rightarrow\hept{\begin{cases}x-0,5=-\left(x+0,4\right)\\x-0,5=x+0,4\end{cases}}\) => x không tồn tại ( ở đay có chút sơ suất ngoặc nhọn đổi thành ngoặc vuông)
b, \(\frac{5}{6}.|\frac{3}{8}-x|-\left(\frac{-7}{8}+\frac{11}{12}-\frac{5}{6}\right)=1\)
,<=> \(|\frac{3}{8}-x|-\left(\frac{-7}{8}+\frac{1}{12}\right)=\frac{6}{5}\)
<=>\(|\frac{3}{8}-x|-\frac{-19}{24}=\frac{6}{5}\)
<=>\(|\frac{3}{8}-x|=\frac{49}{120}\)
=>\(\orbr{\begin{cases}\frac{3}{8}-x=\frac{49}{120}\\\frac{3}{8}-x=\frac{-49}{120}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-1}{30}\\x=\frac{47}{60}\end{cases}}\)
Phần a mình chưa chắc chắn

Câu 1:\(\frac{45^{10}.5^{10}}{75^{10}}\) = \(\frac{\left(5.9\right)^{10}.5^{10}}{\left(5.5.3\right)^{10}}\) = \(\frac{5^{10}.9^{10}.5^{10}}{5^{10}.5^{10}.3^{10}}\) = \(\frac{9^{10}}{3^{10}}\) = \(\frac{3^{10}.3^{10}}{3^{10}}\) = \(3^{10}\) = 59049
Câu 2:\(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}\) = \(\frac{\left(0,4.2\right)^5}{\left(0,4\right)^6}\) = \(\frac{\left(0,4\right)^5.2^5}{\left(0,4\right)^6}\) = \(\frac{2^5}{0,4}\) = \(\frac{32}{0,4}\) = 80
Câu 3:\(\frac{2^{15}.9^4}{6^3.8^3}\) = \(\frac{2^{15}.3^8}{2^{12}.3^3}\) = \(\frac{2^3.3^5}{1.1}\) = \(\frac{8.243}{1}\) = 1944
Câu 4: \(\frac{8^{10}+4^{10}}{8^4+4^{11}}\) = \(\frac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}\) = \(\frac{2^{30}+2^{20}}{2^{12}+2^{22}}\) = \(\frac{2^{20}.2^{10}+2^{20}}{2^{12}+2^{12}.2^{10}}\) = \(\frac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}\) = \(\frac{2^{20}}{2^{12}}\) = \(\frac{2^8}{1}\) = \(2^8\) = 256

a, Ta có: \(\frac{0,8^5}{0,4^6}=\frac{\left(0,4.2\right)^5}{0,4^6}=\frac{0,4^5.2^5}{0,4^6}\) \(=\frac{2^5}{0,4}=80\)
b, Ta có: \(\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}\) \(=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}=\frac{2^{12}\left(2^{18}+2^8\right)}{2^{12}\left(1+2^{10}\right)}\)
\(=\frac{2^{18}+2^8}{1+2^{10}}=\frac{2^8\left(2^{10}+1\right)}{2^{10}+1}=2^8\)
c, Ta có: \(\frac{2^{15}.9^4}{6^3.8^3}=\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^3.\left(2^3\right)^3}=\frac{2^{15}.3^8}{2^3.3^3.2^9}\) \(=\frac{2^{15}.3^8}{2^{12}.3^3}=2^3.3^5=1944\)
b)\(\frac{8^{10}+4^{10}}{8^4+4^{11}}\)=\(\frac{\left(2.4\right)^{10}+4^{10}}{\left(2.4\right)^{10}+4^{11}}\)=\(\frac{2^{10}.4^{10}+4^{10}.1}{2^{10}.4^{10}+4^{10}.4}\)=\(\frac{4^{10}\left(2^{10}+1\right)}{4^{10}\left(2^{10}+4\right)}\)=\(\frac{4^{10}.1025}{4^{10}.1028}\)=\(\frac{1025}{1028}\)
pls nhanh làm ơn
\(\frac16\) - 0,4.\(\frac58\)
= \(\frac16\) - \(\frac14\)
= \(\frac{2}{12}-\frac{3}{12}\)
= - \(\frac{1}{12}\)