Question

help immediatly its a 9th class ques

Answer 1

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$\bf{Simplify~:~ \dfrac{x^{a+b}.~x^{b+c}.~x^{c+a}}{(x^a.~x^b.~x^c)}}$

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$\bf {Using ~the~ law: ~x^m÷x^n=x^{m-n}~ for ~ x^{a+b}÷x^a~and~ so~on:}$

$\sf \qquad \dashrightarrow \: { {x}^{(a + b) - a} . \: {x}^{(b + c) - b} . \: {x}^{(c + a) - c} }$

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$\sf \qquad \dashrightarrow \: { {x}^{(a - a + b)} . \: {x}^{(b - b + c)} . \: {x}^{(c - c + a)} }$

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$\sf \qquad \dashrightarrow \: {x}^{b} . \: {x}^{c} . \: {x}^{a}$

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$\boxed{\bf{\therefore~The~ value~ of~\dfrac{x^{a+b}.~ x^{b+c}.~x^{c+a}}{x^a.~ x^b.~ x^c}~ is ~ x^b.~ x^c.~ x^a.}}$

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$\bf{ Some~ important~ laws~ of~ exponents:}$

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$\sf \qquad \dashrightarrow \: {a}^{m} . \: {a}^{n} = {a}^{m + n}$

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$\sf \qquad \dashrightarrow \: {a}^{m} \div {a}^{n} = {a}^{m - n}$

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$\sf \qquad \dashrightarrow \: ( {a ^{m} )}^{n} = {a}^{mn}$

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$\sf \qquad \dashrightarrow \: {a}^{m} . \: {b}^{m} = {ab}^{m}$

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$\sf \qquad \dashrightarrow \: {a}^{ - 1} = \dfrac{1}{a}$

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$\sf \qquad \dashrightarrow \: {a}^{0} = 1$

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