Question

\(\dfrac{\left(-1\right)^{2018}\cdot\left(\dfrac{2}{5}\right)^3\cdot\left(\dfrac{15}{4}\right)^2}{\dfrac{15^2}{2^{\text{4}}}\cdot\left(\dfrac{2}{5}\right)^3}\)

Answer 1

\(\dfrac{\left(-1\right)^{2018}.\left(\dfrac{2}{5}\right)^3.\left(\dfrac{15}{4}\right)^2}{\dfrac{15^2}{2^4}.\left(\dfrac{2}{5}\right)^3}=\dfrac{1.\left(\dfrac{15}{2^2}\right)^2}{\dfrac{15^2}{2^4}}=\dfrac{15}{2^4}:\dfrac{15}{2^4}=1\)

Answer 2

\(\dfrac{\left(-1\right)^{2018}\cdot\left(\dfrac{2}{5}\right)^3\cdot\left(\dfrac{15}{4}\right)^2}{\dfrac{15^2}{2^4}\cdot\left(\dfrac{2}{5}\right)^3}\)

\(=\dfrac{1\cdot\left(\dfrac{2}{5}\right)^3\cdot\left(\dfrac{15}{4}\right)^2}{\dfrac{15^2}{2^4}\cdot\left(\dfrac{2}{5}\right)^3}\)

\(=\dfrac{1\cdot\left(\dfrac{15}{4}\right)^2}{\dfrac{15^2}{2^4}}\)

\(=\dfrac{\left(\dfrac{15}{4}\right)^2}{\dfrac{15^2}{2^4}}\)

\(=\dfrac{\dfrac{15^2}{4^2}}{\dfrac{15^2}{2^4}}\)

\(=\dfrac{4^2}{2^4}=\dfrac{16}{16}=1\)