Math, asked by babu7689, 11 months ago

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

Answers

Answered by Geethikachandra
0

Answer:

The answer can be 0.00409

Answered by Salmonpanna2022
1

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...☺

:)

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