Question

Simplify: 9-3/4 ÷[2-1/6 + {4-1/3 - (1-1/2 + 1-3/4) } ]​

Answer 1

We have:

$9\dfrac{3}{4}$ ÷ $[2\dfrac{1}{6} +$ {$4\dfrac{1}{3} +(1\dfrac{1}{2}+1\dfrac{3}{4})$}]

We have to simply $9\dfrac{3}{4}$ ÷ $[2\dfrac{1}{6} +$ {$4\dfrac{1}{3} +(1\dfrac{1}{2}+1\dfrac{3}{4})$}].

Solution:

∴ $9\dfrac{3}{4}$ ÷ $[2\dfrac{1}{6} +$ {$4\dfrac{1}{3} +(1\dfrac{1}{2}+1\dfrac{3}{4})$}]

= $\dfrac{39}{4}$ ÷ $[\dfrac{13}{6} +$ {$\dfrac{13}{3} +(\dfrac{3}{2}+\dfrac{7}{4})$}]

= $\dfrac{39}{4}$ ÷ $[\dfrac{13}{6} +$ {$\dfrac{13}{3} +(\dfrac{6+7}{4})$}]

= $\dfrac{39}{4}$ ÷ $[\dfrac{13}{6} +$ {$\dfrac{13}{3} +\dfrac{13}{4}$}]

= $\dfrac{39}{4}$ ÷ $[\dfrac{13}{6} +$ {$\dfrac{26+13}{12}$}]

= $\dfrac{39}{4}$ ÷ $[\dfrac{13}{6} +$ $\dfrac{39}{12}$]

= $\dfrac{39}{4}$ ÷  $\dfrac{39}{12}$

= $\dfrac{39}{4}$  ×  $\dfrac{12}{39}$

= 3

∴ $9\dfrac{3}{4}$ ÷ $[2\dfrac{1}{6} +$ {$4\dfrac{1}{3} +(1\dfrac{1}{2}+1\dfrac{3}{4})$}] = 3

Answer 2

Given:

9-3/4 ÷[2-1/6 + {4-1/3 - (1-1/2 + 1-3/4) } ]​

To find:

The simplified value fo the given expression.

Solution:

Here we have given the numerical expression which is solved by the BODMAS rule.

So we have,

9-3/4 ÷[2-1/6 + {4-1/3 - (1-1/2 + 1-3/4) } ]​

9-3/4 ÷[2-1/6 + {4-1/3 - (2-5/4) } ]​

9-3/4 ÷[2-1/6 + {4-1/3 - 3/4 } ]​

9-3/4 ÷[2-1/6 + {4 - 13/12 } ]​

9-3/4 ÷[2-1/6 +35/12]​

9-3/4 ÷[2 + 33/12]​

9-3/4 ÷57/12

9 - 3/19

168/19

Therefore,

The simplified form of the givne expression is 168/19