What is [((3-1/2)(1+1/4))/((2-2/3)(1/2+1/3))]^2 ?

The solution to [((3-1/2)(1+1/4))/((2-2/3)(1/2+1/3))]^2 is 2025/256

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simplify [((3-1/2)(1+1/4))/((2-2/3)(1/2+1/3))]^2

simplify

[\frac{( 3 - \frac{1}{2} ) ( 1 + \frac{1}{4} )}{( 2 - \frac{2}{3} ) ( \frac{1}{2} + \frac{1}{3} )}]^{2}

\frac{2025}{256}

Decimal Notation

7.91015625

Solution steps

(\frac{( 3 - \frac{1}{2} ) ( 1 + \frac{1}{4} )}{( 2 - \frac{2}{3} ) ( \frac{1}{2} + \frac{1}{3} )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{( ( 3 - \frac{1}{2} ) ( 1 + \frac{1}{4} ) ) ^{2}}{( ( 2 - \frac{2}{3} ) ( \frac{1}{2} + \frac{1}{3} ) ) ^{2}}

((3 - \frac{1}{2}) (1 + \frac{1}{4}))^{2} = \frac{625}{64}

= \frac{\frac{625}{64}}{( ( 2 - \frac{2}{3} ) ( \frac{1}{2} + \frac{1}{3} ) ) ^{2}}

((2 - \frac{2}{3}) (\frac{1}{2} + \frac{1}{3}))^{2} = \frac{100}{81}

= \frac{\frac{625}{64}}{\frac{100}{81}}

\frac{\frac{625}{64}}{\frac{100}{81}} = \frac{2025}{256}

= \frac{2025}{256}

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What is [((3-1/2)(1+1/4))/((2-2/3)(1/2+1/3))]^2 ?
The solution to [((3-1/2)(1+1/4))/((2-2/3)(1/2+1/3))]^2 is 2025/256