Question

[Võ Thạch Đức Tín]

Tính : \(\frac{25^3.3+25^3.5}{2^{27}}\)

Answer 1

[Nobi Nobita]

Đặt \(A=\frac{25^3.3+25^3.5}{2^{27}}\)

   Ta có:\(A=\frac{25^3\left(3+5\right)}{2^{27}}\)

             \(A=\frac{25^3.8}{2^{27}}\)

              \(A=\frac{25^3.2^3}{\left(2^9\right)^3}\)

              \(A=\frac{50^3}{512^3}\)

              \(A=\left(50:512\right)^3\)

              \(A=\left(\frac{25}{256}\right)^3\)

Vậy \(A=\left(\frac{25}{256}\right)^3\)

Answer 2

[Bạch Y]

\(\frac{25^3\cdot3+25^3\cdot5}{2^{27}}=\frac{25^3\left(3+5\right)}{2^{27}}=\frac{5^3\cdot2^3}{2^{27}}=\frac{10^3}{\left(2^9\right)^3}=\left(\frac{5}{256}\right)^3\)