Question

$$( \frac { 3 } { 7 } ) ^ { - 1 } + 0 , 3 ^ { - 3 } + ( - 0 , 5 ) ^ { - 2 } \cdot \frac { 3 } { 4 } - 10 \frac { 19 } { 27 }$$

Answer

INF,75/4

Solution


Use Negative Power Rule: \({x}^{-a}=\frac{1}{{x}^{a}}\).
\[-\frac{1}{{13}^{3}}+{(0,5)}^{-2}\times \frac{3}{4}-\frac{1019}{27}\]
Simplify  \({13}^{3}\)  to  \(2197\).
\[-\frac{1}{2197}+{(0,5)}^{-2}\times \frac{3}{4}-\frac{1019}{27}\]
Use Negative Power Rule: \({x}^{-a}=\frac{1}{{x}^{a}}\).
\[-\frac{1}{2197}+(\frac{1}{0},{5}^{2})\times \frac{3}{4}-\frac{1019}{27}\]
Simplify  \({5}^{2}\)  to  \(25\).
\[-\frac{1}{2197}+(\frac{1}{0},25)\times \frac{3}{4}-\frac{1019}{27}\]
Simplify  \(\frac{1}{0}\)  to  \(\infty \).
\[-\frac{1}{2197}+(\infty ,25)\times \frac{3}{4}-\frac{1019}{27}\]
Simplify  \((\infty ,25)\times \frac{3}{4}\)  to  \(\frac{\infty ,25\times 3}{4}\).
\[-\frac{1}{2197}+\frac{\infty ,25\times 3}{4}-\frac{1019}{27}\]
Simplify  \(25\times 3\)  to  \(75\).
\[-\frac{1}{2197}+\frac{\infty ,75}{4}-\frac{1019}{27}\]
Simplify  \(\frac{\infty ,75}{4}\)  to  \(\infty ,\frac{75}{4}\).
\[-\frac{1}{2197}+(\infty ,\frac{75}{4})-\frac{1019}{27}\]
Collect like terms.
\[(-\frac{1}{2197}-\frac{1019}{27})+\infty ,\frac{75}{4}\]
Simplify  \(-\frac{1}{2197}-\frac{1019}{27}\)  to  \(-\frac{2238770}{59319}\).
\[-\frac{2238770}{59319}+\infty ,\frac{75}{4}\]
Simplify  \(-\frac{2238770}{59319}+\infty \)  to  \(\infty \).
\[\infty ,\frac{75}{4}\]