Question

If \[a = 5\] and \[b = 403\], then find the value of \[\left\{ {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}} + \dfrac{2}{{a + b}}\left( {\dfrac{1}{a} + \dfrac{1}{b}} \right)} \right\}\left[ {{{\left\{ {\dfrac{{{{\left( {a + b} \right)}^2}}}{{ab}}} \right\}}^{ - 1}}} \right]\].

Answer 1

Hint:
Here, we will find the value of the given expression. We will simplify the given expression using rules of exponents and algebraic identities. Then, we will substitute the given values and simplify the expression further to obtain the required answer.

Complete step by step solution:
We will simplify the given expression using algebraic identities and then substitute the given values to get the answer.
First, we will take the L.C.M. of the terms in the parentheses and add them.
Taking the L.C.M. of \[\dfrac{1}{a}\] and \[\dfrac{1}{b}\] in the expression, we get
\[ \Rightarrow \left\{ {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}} + \dfrac{2}{{a + b}}\left( {\dfrac{1}{a} + \dfrac{1}{b}} \right)} \right\}\left[ {{{\left\{ {\dfrac{{{{\left( {a + b} \right)}^2}}}{{ab}}} \right\}}^{ - 1}}} \right] = \left\{ {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}} + \dfrac{2}{{a + b}}\left( {\dfrac{{a + b}}{{ab}}} \right)} \right\}\left[ {{{\left\{ {\dfrac{{{{\left( {a + b} \right)}^2}}}{{ab}}} \right\}}^{ - 1}}} \right]\]
We can cancel out the terms that are common in the numerator and the denominator.
Cancelling the terms, we get
\[ \Rightarrow \left\{ {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}} + \dfrac{2}{{a + b}}\left( {\dfrac{1}{a} + \dfrac{1}{b}} \right)} \right\}\left[ {{{\left\{ {\dfrac{{{{\left( {a + b} \right)}^2}}}{{ab}}} \right\}}^{ - 1}}} \right] = \left\{ {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}} + \dfrac{2}{{ab}}} \right\}\left[ {{{\left\{ {\dfrac{{{{\left( {a + b} \right)}^2}}}{{ab}}} \right\}}^{ - 1}}} \right]\]
We know that \[{\left( {\dfrac{a}{b}} \right)^{ - 1}} = \dfrac{b}{a}\].
Therefore, we can simplify the expression to get
\[ \Rightarrow \left\{ {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}} + \dfrac{2}{{a + b}}\left( {\dfrac{1}{a} + \dfrac{1}{b}} \right)} \right\}\left[ {{{\left\{ {\dfrac{{{{\left( {a + b} \right)}^2}}}{{ab}}} \right\}}^{ - 1}}} \right] = \left\{ {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}} + \dfrac{2}{{ab}}} \right\}\left\{ {\dfrac{{ab}}{{{{\left( {a + b} \right)}^2}}}} \right\}\]
Now, we can rewrite the expression as
\[ \Rightarrow \left\{ {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}} + \dfrac{2}{{a + b}}\left( {\dfrac{1}{a} + \dfrac{1}{b}} \right)} \right\}\left[ {{{\left\{ {\dfrac{{{{\left( {a + b} \right)}^2}}}{{ab}}} \right\}}^{ - 1}}} \right] = \left\{ {{{\left( {\dfrac{1}{a}} \right)}^2} + {{\left( {\dfrac{1}{b}} \right)}^2} + 2\left( {\dfrac{1}{a}} \right)\left( {\dfrac{1}{b}} \right)} \right\}\left\{ {\dfrac{{ab}}{{{{\left( {a + b} \right)}^2}}}} \right\}\]
We know that the square of the sum of two numbers is given by the algebraic identity \[{\left( {x + y} \right)^2} = {x^2} + {y^2} + 2xy\].
Substituting \[x = \dfrac{1}{a}\] and \[y = \dfrac{1}{b}\] in the identity, we get
\[{\left( {\dfrac{1}{a} + \dfrac{1}{b}} \right)^2} = {\left( {\dfrac{1}{a}} \right)^2} + {\left( {\dfrac{1}{b}} \right)^2} + 2\left( {\dfrac{1}{a}} \right)\left( {\dfrac{1}{b}} \right)\]
Next, we will substitute \[{\left( {\dfrac{1}{a}} \right)^2} + {\left( {\dfrac{1}{b}} \right)^2} + 2\left( {\dfrac{1}{a}} \right)\left( {\dfrac{1}{b}} \right) = {\left( {\dfrac{1}{a} + \dfrac{1}{b}} \right)^2}\] in the expression \[\left\{ {{{\left( {\dfrac{1}{a}} \right)}^2} + {{\left( {\dfrac{1}{b}} \right)}^2} + 2\left( {\dfrac{1}{a}} \right)\left( {\dfrac{1}{b}} \right)} \right\}\left\{ {\dfrac{{ab}}{{{{\left( {a + b} \right)}^2}}}} \right\}\].
Thus, we get
\[ \Rightarrow \left\{ {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}} + \dfrac{2}{{a + b}}\left( {\dfrac{1}{a} + \dfrac{1}{b}} \right)} \right\}\left[ {{{\left\{ {\dfrac{{{{\left( {a + b} \right)}^2}}}{{ab}}} \right\}}^{ - 1}}} \right] = \left\{ {{{\left( {\dfrac{1}{a} + \dfrac{1}{b}} \right)}^2}} \right\}\left\{ {\dfrac{{ab}}{{{{\left( {a + b} \right)}^2}}}} \right\}\]
Taking the L.C.M. of \[\dfrac{1}{a}\] and \[\dfrac{1}{b}\] in the expression, we get
\[ \Rightarrow \left\{ {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}} + \dfrac{2}{{a + b}}\left( {\dfrac{1}{a} + \dfrac{1}{b}} \right)} \right\}\left[ {{{\left\{ {\dfrac{{{{\left( {a + b} \right)}^2}}}{{ab}}} \right\}}^{ - 1}}} \right] = \left\{ {{{\left( {\dfrac{{a + b}}{{ab}}} \right)}^2}} \right\}\left\{ {\dfrac{{ab}}{{{{\left( {a + b} \right)}^2}}}} \right\}\]
We can rewrite the expression using the rule of exponents \[{\left( {\dfrac{a}{b}} \right)^m} = \dfrac{{{a^m}}}{{{b^m}}}\].
Therefore, we get
\[ \Rightarrow \left\{ {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}} + \dfrac{2}{{a + b}}\left( {\dfrac{1}{a} + \dfrac{1}{b}} \right)} \right\}\left[ {{{\left\{ {\dfrac{{{{\left( {a + b} \right)}^2}}}{{ab}}} \right\}}^{ - 1}}} \right] = \left\{ {\dfrac{{{{\left( {a + b} \right)}^2}}}{{{{\left( {ab} \right)}^2}}}} \right\}\left\{ {\dfrac{{ab}}{{{{\left( {a + b} \right)}^2}}}} \right\}\]
We can cancel out the terms that are common in the numerator and the denominator.
Cancelling the terms, we get
\[ \Rightarrow \left\{ {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}} + \dfrac{2}{{a + b}}\left( {\dfrac{1}{a} + \dfrac{1}{b}} \right)} \right\}\left[ {{{\left\{ {\dfrac{{{{\left( {a + b} \right)}^2}}}{{ab}}} \right\}}^{ - 1}}} \right] = \dfrac{1}{{ab}}\]
Therefore, we have simplified the expression to obtain a much simpler expression.
We will substitute \[a = 5\] and \[b = 403\] to get the required answer.
Substituting \[a = 5\] and \[b = 403\] in the expression, we get
\[ \Rightarrow \left\{ {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}} + \dfrac{2}{{a + b}}\left( {\dfrac{1}{a} + \dfrac{1}{b}} \right)} \right\}\left[ {{{\left\{ {\dfrac{{{{\left( {a + b} \right)}^2}}}{{ab}}} \right\}}^{ - 1}}} \right] = \dfrac{1}{{5 \times 403}}\]
Multiplying 5 by 403, we get the value of the expression as

\[\therefore\] We get the value of the expression \[\left\{ {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}} + \dfrac{2}{{a + b}}\left( {\dfrac{1}{a} + \dfrac{1}{b}} \right)} \right\}\left[ {{{\left\{ {\dfrac{{{{\left( {a + b} \right)}^2}}}{{ab}}} \right\}}^{ - 1}}} \right]\] as \[\dfrac{1}{{2015}}\].

Note:
It is important for us to remember the rules of exponents are \[{\left( {\dfrac{a}{b}} \right)^{ - 1}} = \dfrac{b}{a}\] and \[{\left( {\dfrac{a}{b}} \right)^m} = \dfrac{{{a^m}}}{{{b^m}}}\]. Also, the square of the sum of two numbers is given by the algebraic identity \[{\left( {x + y} \right)^2} = {x^2} + {y^2} + 2xy\]. We can also solve the question by directly substituting the values in the given expression but it will include much larger numbers. Thus, making it difficult for us to calculate. For example, we will need to take the L.C.M. of \[{5^2}\], \[{403^2}\], and \[403 \times 5\] to solve the first parentheses.