Question

pls help me
it is hard to understand
if you solve this answer I will give 50 points to you​

Answer 1

Question :-

$\sf\dfrac{ {3}^{ - 3} \times {6}^{2} \times \sqrt{98} }{ {5}^{2} \times \sqrt[3]{ \frac{1}{25}} \times {15}^{ \frac{ - 4}{3} } \times {3}^{\frac{1}{3} }}$

Answer :-

LHS :-

$\implies\sf \dfrac{ {3}^{ - 3} \times {6}^{2} \times \sqrt{98} }{ {5}^{2} \times \sqrt[3]{ \frac{1}{25}} \times {15}^{ \frac{ - 4}{3} } \times {3}^{ \frac{1}{3} } }$

$\implies\sf\dfrac{ {3}^{ - 3} \times {(3 \times 2)}^{2} \times \sqrt{7 \times 7 \times 2} }{ {5}^{2} \times \sqrt[3]{ \frac{1}{ {5}^{2} }} \times {(5 \times 3)}^{ \frac{ - 4}{3} } \times {3}^{ \frac{1}{3}}}$

$\implies\sf \dfrac{ {3}^{ - 3} \times {3}^{2} \times {2}^{2} \times 7 \sqrt{2} }{ {5}^{2} \times \frac{1}{ {5}^{ \frac{2}{3} } } \times {5}^{ \frac{ - 4}{3} } \times 3 ^\frac{1}{3} }$

$\implies\sf \dfrac{ {3}^{ - 3 + 2} \times {2}^{2} \times 7 \times {2}^{ \frac{1}{2} } }{ {5}^{2} \times {5}^{ \frac{ - 2}{3} } \times {5}^{ \frac{ - 4}{3} } \times {3}^{ \frac{ - 4}{3} + \frac{1}{3} } }$

$\implies\sf \dfrac{ {3}^{ - 1} \times 4 \times \sqrt{2} \times 7 }{ {5}^{2 - \frac{2}{3} - \frac{4}{3} } \times {3}^{ \frac{ - 3}{3} } }$

$\implies\sf \dfrac{ \cancel{{3}^{ - 1}} \times 4 \times \sqrt{2} \times 7 }{ {5}^{2 - \frac{6}{3} } \times \cancel{ {3}^{ - 1} }}$

$\implies\sf \dfrac{28 \sqrt{2} }{ {5}^{2 - 2} }$

$\implies\sf \dfrac{28 \sqrt{2} }{ {5}^{0} }$

$\implies\sf 28 \sqrt{2}$

ㅤ

$\sf RHS = 28 \sqrt{2}$

ㅤ

LHS = RHS

$\begin{gathered}\boxed{\begin{minipage}{5 cm}\bf{\dag}\:\:\underline{\text{Law of Exponents :}}\\\\\bigstar\:\:\sf\dfrac{a^m}{a^n} = a^{m - n}\\\\\bigstar\:\:\sf{(a^m)^n = a^{mn}}\\\\\bigstar\:\:\sf(a^m)(a^n) = a^{m + n}\\\\\bigstar\:\:\sf\dfrac{1}{a^n} = a^{-n}\\\\\bigstar\:\:\sf\sqrt[\sf n]{\sf a} = (a)^{\dfrac{1}{n}}\end{minipage}}\end{gathered}$