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\(A=\frac{2\cdot\left(2^3\right)^4\cdot\left(3^3\right)^8+2^2\cdot\left(2\cdot3\right)^9}{2^7\cdot6^7+2^7\cdot\left(2^3\cdot5\right).\left(3^2\right)^4}\)
\(A=\frac{2\cdot2^{12}\cdot3^{24}+2^2\cdot\left(2\cdot3\right)^9}{12^7+2^7\cdot\left(2^3\cdot5\right)\cdot3^8}\)
Đến đó thì bí
a: \(=\dfrac{-3^4\cdot2^8}{2^2\cdot2^2\cdot3^2}=-3^2\cdot2^2=-6^2=-36\)
b: \(=\dfrac{3^6\cdot2^{15}}{3^6\cdot2^6\cdot2^{15}}=\dfrac{1}{2^6}=\dfrac{1}{64}\)
c: \(=\left(\dfrac{0.8}{0.4}\right)^5\cdot\dfrac{1}{0.4}=2^5\cdot\dfrac{1}{0.4}=\dfrac{32}{0.4}=80\)
a) \(\left(\frac{1}{3}-\frac{1}{5}\right)^2:\left(\frac{1}{5}\right)^2=\left[\left(\frac{1}{3}-\frac{1}{5}\right):\frac{1}{5}\right]^2=\left(\frac{2}{15}:\frac{1}{5}\right)^2=\left(\frac{2}{3}\right)^2=\frac{4}{9}\)
c)\(7\frac{1}{23}+\frac{10}{27}-5\frac{1}{23}+\frac{17}{27}+2^3=\left(7\frac{1}{23}-5\frac{1}{23}\right)+\left(\frac{10}{27}+\frac{17}{27}\right)+2^3=2+1+8=11\)
d)\(5.\left(-\frac{5}{2}\right)^2+\frac{1}{5}.\left(-3\right)^2=5.\frac{25}{4}+\frac{1}{5}.9=\frac{125}{4}+\frac{9}{5}=\frac{661}{20}\)
\(\frac{2^5,9^3}{27^2.8^2}\)
\(=\frac{2^5.3^5}{3^5.2^5}\)
\(=\frac{1}{1}\)
\(=1\)
= 1 nhé !
Chúc bạn học tốt ^-^