give step by step explanation.
Answers
Answer:
Main idea
We know that,
\rm\cdots\longrightarrow (a^{m})^{n}=a^{mn}\ \cdots[1]⋯⟶(am)n=amn ⋯[1]
We know that,
\rm\cdots\longrightarrow a^{(m^{n})}=a^{m^{n}} \cdots[2]⋯⟶a(mn)=amn⋯[2]
We know that,
\rm\cdots\longrightarrow\sqrt[\rm n]{\rm a^{\rm m}}=a^{\frac{m}{n}}\ \cdots [3]⋯⟶nam=anm ⋯[3]
We know that,
\rm\cdots\longrightarrow(ab)^{n}=a^{n}b^{n}\ \cdots[4]⋯⟶(ab)n=anbn ⋯[4]
Provided that \rm a > 0a>0 and \rm b > 0b>0 .
\rm\large\underline{\text{Explanation}}Explanation
The required expression is the following.
\rm\cdots\longrightarrow\sqrt{\dfrac{m^{n^{2}}n^{m^{2}}a^{(m+n)}}{(m+n)^{(m+n)^{2}}}}⋯⟶(m+n)(m+n)2mn2nm2a(m+n)
According to \rm[2][2] ,
\rm\cdots\longrightarrow\sqrt{\dfrac{m^{(n^{2})}n^{(m^{2})}a^{(m+n)}}{(m+n)^{(m+n)^{2}}}}⋯⟶(m+n)(m+n)2m(n2)n(m2)a(m+n)
According to \rm[3][3] ,
\rm\cdots\longrightarrow\left(\dfrac{m^{(n^{2})}n^{(m^{2})}a^{(m+n)}}{(m+n)^{(m+n)^{2}}}\right)^{\frac{1}{2}}⋯⟶((m+n)(m+n)2m(n2)n(m2)a(m+n))21
According to \rm[1][1] and \rm[4][4] ,
\rm\cdots\longrightarrow\dfrac{m^{\frac{n^{2}}{2}}n^{\frac{m^{2}}{2}}a^{\frac{(m+n)}{2}}}{(m+n)^{\frac{(m+n)^{2}}{2}}}⋯⟶(m+n)2(m+n)2m2n2n2m2a2(m+n)
The correct answer is choice (b) \rm\dfrac{m^{\frac{n^{2}}{2}}n^{\frac{m^{2}}{2}}a^{\frac{m+n}{2}}}{(m+n)^{\frac{(m+n)^{2}}{2}}}(m+n)2(m+n)2m2n2n2m2a2m+n .
\large\textrm{\underline{Extra information}}Extra information
\boxed{\textrm{Zero and negative exponents}}Zero and negative exponents
\rm\cdots\longrightarrow a^{0}=1⋯⟶a0=1
\rm\cdots\longrightarrow a^{-n}=\dfrac{1}{a^{n}}⋯⟶a−n=an1
Provided that