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A.1

B.2

C.4

D.3

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Formula Used:

In this question, we are going to use the binomial theorem, which is:

\[{\left( {x + y} \right)^n} = \sum\limits_{k = 0}^n {\left( \begin{array}{l}n\\k\end{array} \right){x^{n - k}}{y^k}} = \sum\limits_{k = 0}^n {\left( \begin{array}{l}n\\k\end{array} \right){x^k}{y^{n - k}}} \]

Now, the given expression \[x = \dfrac{{1 + 4.343 + 7.4 + 2.3.49 + 7.343}}{{16 + {2^6}{{.3}^1} + {2^5}{{.3}^3} + {2^6}{{.3}^3} + {2^4}{{.3}^4}}}\] can be rearranged as:

\[x = \dfrac{{1 + 7.4 + 2.3.49 + 4.343 + 7.343}}{{16 + {2^6}{{.3}^1} + {2^5}{{.3}^3} + {2^6}{{.3}^3} + {2^4}{{.3}^4}}}\]

Now, by shifting the powers (for e.g., \[{2^3}{.3^5}\] can be written as \[{6^3}{.3^2}\] ), we have:

\[x = \dfrac{{{{1.1}^4}{{.7}^0} + {{4.1}^3}{{.7}^1} + {{6.1}^2}{{.7}^2} + {{4.1}^1}{{.7}^3} + {{1.1}^0}{{.7}^4}}}{{{{1.2}^4}{{.6}^0} + {{4.2}^3}{{.6}^1} + {{6.2}^2}{{.6}^2} + {{4.2}^1}{{.6}^3} + {{1.2}^0}{{.6}^4}}}\]

Now, clearly the above expression makes more sense and can be written as:

\[x = \dfrac{{^4{C_0}{{.1}^4}{{.7}^0}{ + ^4}{C_1}{{.1}^3}{{.7}^1}{ + ^4}{C_2}{{.1}^2}{{.7}^2}{ + ^4}{C_3}{{.1}^1}{{.7}^3}{ + ^4}{C_4}{{.1}^0}{{.7}^4}}}{{^4{C_0}{{.2}^4}{{.6}^0}{ + ^4}{C_1}{{.2}^3}{{.6}^1}{ + ^4}{C_2}{{.2}^2}{{.6}^2}{ + ^4}{C_3}{{.2}^1}{{.6}^3}{ + ^4}{C_4}{{.2}^0}{{.6}^4}}}\]

w.k.t

\[{\left( {x + y} \right)^n} = \sum\limits_{k = 0}^n {\left( \begin{array}{l}n\\k\end{array} \right){x^{n - k}}{y^k}} = \sum\limits_{k = 0}^n {\left( \begin{array}{l}n\\k\end{array} \right){x^k}{y^{n - k}}} \]

Comparing the two equations and now, this expression can be simplified into:

\[x = \dfrac{{{{\left( {1 + 7} \right)}^4}}}{{{{\left( {2 + 6} \right)}^4}}} = \dfrac{{{8^4}}}{{{8^4}}} = 1\]

Hence, the answer of the given expression is \[1\] .

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