Question

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Answer 1

5.

\(\left(\dfrac{1}{3}-\dfrac{1}{2}\right)^2+\left(\dfrac{\sqrt{16}}{119}-3\right)^0-\dfrac{7}{36}\)

\(=\left(\dfrac{2}{6}-\dfrac{3}{6}\right)^2+1-\dfrac{7}{36}\)

\(=\left(-\dfrac{1}{6}\right)^2-\dfrac{7}{36}+1\)

\(=\dfrac{1}{36}-\dfrac{7}{36}+1\)

\(=\dfrac{-6}{36}+1\)

\(=\dfrac{-1}{6}+1=\dfrac{5}{6}\)

6.

\(\dfrac{9^{14}}{3^{12}.27^6}=\dfrac{\left(3^2\right)^{14}}{3^{12}.\left(3^3\right)^6}=\dfrac{3^{28}}{3^{12}.3^{18}}=\dfrac{3^{28}}{3^{30}}=\dfrac{1}{3^2}=\dfrac{1}{9}\)