Question

Simplify (81/16)^-3/4 x [(25/9)^-3/2 / (5/2)^-3]

Answer 1

Answer is should be equal to 1

first write the first term in the term in the form of squares of 9 and 4 this should cancel out -3/4 to -3/2

then take common -3/2 and this cancels out 9 and leaves us with (25/4)^-3/2

Now split -3/2 into 1/2×-3

anything to the power 1/2 is equal to the root of that number that leaves us with (5/2)^-3

now transpose (5/3)^-3 to the numerator and this makes it (5/3)^3

as bases are same the powers get added this leaves us with (5/2)^0 this makes it one

Answer 2

Step-by-step explanation:

$\bf \underline{Given-}$

$\sf{\bigg( \frac{81}{16} \bigg) ^{ - \frac{ 3}{4} } \times \left \{ \bigg(\frac{25}{9} \bigg) ^{ - \frac{3}{2} } \div \bigg( \frac{5}{2} \bigg)^{ - 3} \right \}}$

$\bf \underline{To\: find-}$

$\textsf{Simplify the given fractional expression and find their value.}$

$\bf \underline{Solution-}$

$\textsf{We have,}$

$\sf{\bigg( \frac{81}{16} \bigg) ^{ - \frac{ 3}{4} } \times \left \{ \bigg(\frac{25}{9} \bigg) ^{ - \frac{3}{2} } \div \bigg( \frac{5}{2} \bigg)^{ - 3} \right \}}$

$\Rightarrow{ \sf{\bigg( \frac{16}{81} \bigg) ^{ \frac{ 3}{4} } \times \left \{ \bigg(\frac{9}{25} \bigg) ^{ \frac{3}{2} } \div \bigg( \frac{2}{5} \bigg)^{ 3} \right \}} }$

$\Rightarrow{ \sf{\bigg( \frac{ {2}^{4} }{ {3}^{4} } \bigg) ^{ \frac{ 3}{4} } \times \left \{ \bigg(\frac{ {3}^{2} }{ {5}^{2} } \bigg) ^{ \frac{3}{2} } \div \bigg( \frac{2}{5} \bigg)^{ 3} \right \}} }$

$\Rightarrow{ \sf{\bigg( \frac{2}{3} \bigg) ^{ \cancel{4} \times \frac{ 3}{ \cancel{4}} } \times \left \{ \bigg(\frac{3}{5} \bigg) ^{ \cancel{2} \times \frac{3}{ \cancel{2}} } \div \bigg( \frac{2}{5} \bigg)^{ 3} \right \}} }$

$\Rightarrow{ \sf{\bigg( \frac{2}{3} \bigg) ^{ 3} \times \left \{ \bigg(\frac{3}{5} \bigg) ^{ 3 } \div \bigg( \frac{2}{5} \bigg)^{ 3} \right \}} }$

$\Rightarrow{ \sf{\bigg( \frac{2}{3} \bigg) ^{ 3} \times \left \{ \bigg(\frac{3}{5} \bigg) ^{ 3 } \times \bigg( \frac{5}{2} \bigg)^{ 3} \right \}} }$

$\Rightarrow{ \sf{\bigg( \frac{2}{3} \bigg) ^{ 3} \times } \bigg(\frac{ {3}^{3} }{ \cancel{{5}^{3}} } \times \frac{ \cancel{{5}^{3}} }{ {2}^{3} } \bigg) }$

$\Rightarrow{ \sf{\bigg( \frac{2}{3} \bigg) ^{ 3} \times } \bigg(\frac{ {3}^{3} }{ {2}^{3}} \bigg) }$

$\Rightarrow{ \sf{\bigg( \frac{2}{3} \bigg) ^{ 3} \times } \bigg(\frac{ {3}}{ {2}} \bigg) ^{3} }$

$\Rightarrow{ \sf \frac{ \cancel{ {2}^{3} }}{ \cancel{{3}^{3} }} \times \frac{ \cancel{{3}^{3}} }{ \cancel{ {2}^{3} }} }$

$\Rightarrow{ \sf1}$

$\bf \underline{Hence, after \:simplifying\: the\: value\: of\: \bigg( \frac{81}{16} \bigg) ^{ - \frac{ 3}{4} } \times \left \{ \bigg(\frac{25}{9}\bigg) ^{ - \frac{3}{2} } \div \bigg( \frac{5}{2} \bigg)^{ - 3} \right \} \: is \: 1.}$

$\bf \underline{More\: information:-}$

$\textsf{Low of Integral Exponents}$

For any two real numbers a and b, a, b ≠ 0, and for any two positive integers, m and n

➲ If a be any non - zero rational number, then

a^0 = 1

➲ If a be any non - zero rational number and m,n be integer, then

(a^m)^n = a^mn

➲ If a be any non - zero rational number and m be any positive integer, then

a^-m = 1/a^m

➲ If a/b is a rational number and m is a positive integer, then

(a/b)^m = a^m/b^m

➲ For any Integers m and n and any rational number a, a ≠ 0

a^m × a^n = a^m+n

➲ For any Integers m and n for non - zero rational number a,

a^m ÷ a^n = a^m-n

➲ If a and b are non - zero rational numbers and m is any integer, then

(a+b)^m = a^m × b^m

I hope it's help you...☺

:)