Simplify the given expression.
\[{{\left[ {{\left( 4 \right)}^{-1}}-\left( 5 \right)-1 \right]}^{2}}\times {{\left( \dfrac{5}{8} \right)}^{-1}}\]
Answer
681.3k+ views
Hint: Find the multiplicative inverse of the function. Undergo algebraic operations like subtraction and multiplication, thus simplify the given expression.
Complete step-by-step answer:
Given is the expression \[{{\left[ {{\left( 4 \right)}^{-1}}-\left( 5 \right)-1 \right]}^{2}}\times {{\left( \dfrac{5}{8} \right)}^{-1}}\].
A multiplicative inverse or reciprocal for a number x is denoted by \[\dfrac{1}{x}\] or \[{{x}^{-1}}\] is a number which when multiplied by x yields the multiplicative identity 1.
The multiplicative inverse of a fraction \[\dfrac{a}{b}\] is \[\dfrac{b}{a}\]. For the multiplicative inverse of a real number, divide 1 by the number.
We have been given, \[{{\left[ {{\left( 4 \right)}^{-1}}-\left( 5 \right)-1 \right]}^{2}}\times {{\left( \dfrac{5}{8} \right)}^{-1}}..........(1)\]
\[{{4}^{-1}}={\dfrac{1}{4}},{{5}^{-1}}={\dfrac{1}{5}},{{\left( \dfrac{5}{8} \right)}^{-1}}=\dfrac{8}{5}\], these are the multiplicative inverse of the given real number. Now let us substitute the values in equation (1).
\[{{\left[ {{\left( 4 \right)}^{-1}}-\left( 5 \right)-1 \right]}^{2}}\times {{\left( \dfrac{5}{8} \right)}^{-1}}={{\left( \dfrac{1}{4}-\dfrac{1}{5} \right)}^{2}}\times \left( \dfrac{8}{5} \right)\]
Now let us simplify the expression, we got,
\[\begin{align}
& ={{\left( \dfrac{5-4}{4\times 5} \right)}^{2}}\times \left( \dfrac{8}{5} \right)={{\left( \dfrac{1}{20} \right)}^{2}}\times \dfrac{8}{5} \\
& =\dfrac{1}{20}\times \dfrac{1}{20}\times \dfrac{8}{5}=\dfrac{1}{20}\times \dfrac{2}{5}\times \dfrac{1}{5} \\
& =\dfrac{1}{10\times 5\times 5}=\dfrac{1}{10\times 25}=\dfrac{1}{250}=0.004 \\
\end{align}\]
Thus we got the value of,
\[{{\left[ {{\left( 4 \right)}^{-1}}-\left( 5 \right)-1 \right]}^{2}}\times {{\left( \dfrac{5}{8} \right)}^{-1}}=\dfrac{1}{250}=0.004\]
\[\therefore \]We have simplified the given expression.
Note: In the phrase multiplicative inverse, the qualifier multiplicative is after committed and then tacitly understood. Multiplicative inverse can be defined over many mathematical domains as well as numbers. In case of real numbers, zero does not have a reciprocal because no real numbers multiplied by 0 produce 1.
Complete step-by-step answer:
Given is the expression \[{{\left[ {{\left( 4 \right)}^{-1}}-\left( 5 \right)-1 \right]}^{2}}\times {{\left( \dfrac{5}{8} \right)}^{-1}}\].
A multiplicative inverse or reciprocal for a number x is denoted by \[\dfrac{1}{x}\] or \[{{x}^{-1}}\] is a number which when multiplied by x yields the multiplicative identity 1.
The multiplicative inverse of a fraction \[\dfrac{a}{b}\] is \[\dfrac{b}{a}\]. For the multiplicative inverse of a real number, divide 1 by the number.
We have been given, \[{{\left[ {{\left( 4 \right)}^{-1}}-\left( 5 \right)-1 \right]}^{2}}\times {{\left( \dfrac{5}{8} \right)}^{-1}}..........(1)\]
\[{{4}^{-1}}={\dfrac{1}{4}},{{5}^{-1}}={\dfrac{1}{5}},{{\left( \dfrac{5}{8} \right)}^{-1}}=\dfrac{8}{5}\], these are the multiplicative inverse of the given real number. Now let us substitute the values in equation (1).
\[{{\left[ {{\left( 4 \right)}^{-1}}-\left( 5 \right)-1 \right]}^{2}}\times {{\left( \dfrac{5}{8} \right)}^{-1}}={{\left( \dfrac{1}{4}-\dfrac{1}{5} \right)}^{2}}\times \left( \dfrac{8}{5} \right)\]
Now let us simplify the expression, we got,
\[\begin{align}
& ={{\left( \dfrac{5-4}{4\times 5} \right)}^{2}}\times \left( \dfrac{8}{5} \right)={{\left( \dfrac{1}{20} \right)}^{2}}\times \dfrac{8}{5} \\
& =\dfrac{1}{20}\times \dfrac{1}{20}\times \dfrac{8}{5}=\dfrac{1}{20}\times \dfrac{2}{5}\times \dfrac{1}{5} \\
& =\dfrac{1}{10\times 5\times 5}=\dfrac{1}{10\times 25}=\dfrac{1}{250}=0.004 \\
\end{align}\]
Thus we got the value of,
\[{{\left[ {{\left( 4 \right)}^{-1}}-\left( 5 \right)-1 \right]}^{2}}\times {{\left( \dfrac{5}{8} \right)}^{-1}}=\dfrac{1}{250}=0.004\]
\[\therefore \]We have simplified the given expression.
Note: In the phrase multiplicative inverse, the qualifier multiplicative is after committed and then tacitly understood. Multiplicative inverse can be defined over many mathematical domains as well as numbers. In case of real numbers, zero does not have a reciprocal because no real numbers multiplied by 0 produce 1.
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