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(A) To solve the expression,we follow the $BODMAS$ rule (Bracket,Of,Division,Multiplication,Addition,Subtraction).First,solve the vinculum (bar bracke

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(A) To solve the expression,we follow the $BODMAS$ rule (Bracket,Of,Division,Multiplication,Addition,Subtraction).First,solve the vinculum (bar bracket): $\frac{2}{3} + \frac{1}{4} = \frac{8+3}{12} = \frac{11}{12}$.Next,solve the square root: $\sqrt{\frac{144}{121}} = \frac{12}{11}$.Substitute these into the inner bracket: $\left( \frac{11}{12} 	imes \frac{12}{11} + 23 ight) = (1 + 23) = 24$.Now,substitute this back into the curly bracket: $\{ 4 	imes 24 \div 12 + 5 \}$.Perform division and multiplication: $4 	imes (24 \div 12) + 5 = 4 	imes 2 + 5 = 8 + 5 = 13$.Finally,substitute into the main expression: $[10 + 13 - 3] = 20$.

$\left[ {10 + \left\{ {4 \times \left( {\overline {\frac{2}{3} + \frac{1}{4}} \times \sqrt {\frac{{144}}{{121}}} + 23} \right) \div 12 + 5} \right\} - 3} \right] = ?$

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