Question

Class 9 mathematics (Question in attached image)

Answer 1

Answer:

Follow  i n s t a g r a m = niral161

Step-by-step explanation:

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Answer 2

Question:-

$\longrightarrow\sf{ \bigg(\dfrac{5^{-1} \times 7^{2} }{5^{2} \times 7^{-4} }\bigg)^{\frac{7}{2} } \times \bigg(\dfrac{5^{-2} \times 7^{3}}{5^{3} \times 7^{-5} } \bigg)^{\frac{-3}{2} }}$

Let's divide the sum into 2 parts in order to understand the sum.

Let the first fraction be part A and the second fraction part B.

Part A :-

Solving part A,

here we have 2 numbers in the numerator and 2 numbers in the denominator respectively.

$\implies\sf{ \bigg(\dfrac{5^{-1}}{5^{2} } \times \dfrac{7^{2}}{7^{-4}} \bigg)^{\frac{7}{2} }}$

Here by applying the exponent law $\longrightarrow \sf{\frac{a^{m}}{a^{n}} = a^{m-n}}$

$\implies \sf{(5^{-1 - 2} \times 7^{2 - (-4)})^{\frac{7}{2}} }$

$\implies \sf{(5^{-3} \times 7^{6} )^{\frac{7}{2} }}$

Exponent law :- $\longrightarrow \sf{a^{-m} = \dfrac{1}{a^{m}}}$

$\implies \sf{\bigg(\dfrac{7^{6}}{5^{3}} \bigg)^{\frac{7}{2} }}$

Part B :-

$\implies \sf{ \bigg(\dfrac{5^{-2} \times 7^{3}}{5^{3} \times 7^{-5}} \bigg)^{\frac{-3}{2} }}$

Here again we have 2 numbers in the numerator and the denominator. Expressing them as 2 fractions we get :-

$\implies \sf{\bigg(\dfrac{5^{-2}}{5^{3}} \times \dfrac{7^{3}}{7^{-5}} \bigg)^{\frac{-3}{2} }}$

Applying the exponent law again,

$\implies \sf{(5^{-2 - 3} \times 7^{3 - (-5)})^{\frac{-3}{2} }}$

$\implies\sf{ (5^{-5} \times 7^{8})^{ \frac{-3}{2} }}$

Apply exponent law,

$\implies \sf{\bigg(\dfrac{7^{8}}{5^{5}} \bigg)^{\frac{-3}{2} }}$

Let's now write both the parts as one set.

$\longrightarrow \sf{\bigg({\dfrac{7^{6}}{5^{3}} \bigg)^{ \frac{7}{2} }\times \bigg(\dfrac{7^{8}}{5^{5}} \bigg)^{\frac{-3}{2} }}}$

Let's multiply the whole powers with the fraction itself.

• $\longrightarrow \sf{(a^{m})^{n} = a^{mn}}$

$\implies \sf { (7^{6\times\frac{7}{2} } \times 5^{-3 \times \frac{7}{2} }) \times (7^{8 \times \frac{-3}{2} } \times 5^{-5 \times \frac{-3}{2} })}$

Cancelling,

$\implies \sf{(7^{\not6 \times \frac{7}{\not2} } \times 5^{\frac{-21}{2}} ) \times (7^{\not8 \times \frac{-3}{\not2}} \times 5^{ \times \frac{15}{2} })}$

$\implies \sf{ (7^{21} \times 5^{\frac{-21}{2} }) \times (7^{-12} \times 5^{\frac{15}{2} })}$

Removing the brackets,

$\implies 7^{21} \times 5^{\frac{-21}{2} } \times 7^{-12} \times 5^{\frac{15}{2} }$

Grouping,

$\implies \sf{(7^{21} \times 7^{-12}) \times (5^{\frac{-21}{2} } \times 5^{\frac{15}{2} })}$

• Exponent law :- $\longrightarrow \sf{a^{m} \times a^{n} = a^{m+n}}$

Therefore,

$\implies \sf{(7^{21 + -12}) \times (5^{\frac{-21}{2} + \frac{15}{2} })}$

$\implies \sf{(7^{9}) \times (5^{\frac{-6}{2} })}$

$\implies \sf{ (7^{9}) \times (5^{-3})}$

$\implies \sf{\dfrac{7^{6}}{5^{3}}}$

$\implies \sf{ \dfrac{117649}{125} }$

Important points to note :-

• aᵇ × aⁿ = aᵇ⁺ⁿ
• aᵇ ÷ aⁿ = aᵇ⁻ⁿ
• a⁰ = 1
• $\sf{\sqrt[n]{a} = a^{\frac{1}{n} }}$
• $\sf{ a^{-m} = \dfrac{1}{a^{m}}}$
• aᵇ × bᵇ = (ab)ᵇ
• $\sf{a^{m} \div b^{m} = \bigg(\dfrac{a}{b} \bigg)^{m}}$