Question

maths(SM) ch1pg1.35, q7,8

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Answer 1

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Answer 2

$\mathsf{7.\;\;Given :\;\bigg(\dfrac{x^a}{x^b}\bigg)^{a + b} \times \bigg(\dfrac{x^b}{x^c}\bigg)^{b + c} \times \bigg(\dfrac{x^c}{x^a}\bigg)^{c + a}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{\dfrac{a^m}{a^n} = a^{m - n}}}}$

$\mathsf{\implies \dfrac{x^a}{x^b} = x^{a - b}}$

$\mathsf{\implies \dfrac{x^b}{x^c} = x^{b - c}}$

$\mathsf{\implies \dfrac{x^c}{x^a} = x^{c - a}}$

$\mathsf{\implies \big[x^{a - b}\big]^{a + b} \times \big[x^{b - c}\big]^{b + c} \times \big[x^{c - a}\big]^{c + a}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{\big[a^m\big]^n = a^{mn}}}}$

$\mathsf{\implies \big[x\big]^{(a + b)(a - b)} \times \big[x\big]^{(b + c)(b - c)} \times \big[x\big]^{(c + a)(c - a)}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{(p + q)(p - q) = p^2 - q^2}}}$

$\mathsf{\implies \big[x\big]^{a^2 - b^2} \times \big[x\big]^{b^2 - c^2} \times \big[x\big]^{c^2 - a^2}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{a^m \times a^n = a^{m + n}}}}$

$\mathsf{\implies \big[x\big]^{a^2 - b^2 + b^2 - c^2 + c^2 - a^2}}$

$\mathsf{\implies \big[x\big]^{0}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{a^0 = 1}}}$

$\mathsf{\implies \big[x\big]^{0} = 1}$

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$\mathsf{8.\;\;Given :{\sqrt[a + b]{\dfrac{x^{a^{2}}}{x^{b^{2}}}}\;\times\sqrt[b + c]{\dfrac{x^{b^{2}}}{x^{c^{2}}}}\;\times\sqrt[c + a]{\dfrac{x^{c^{2}}}{x^{a^{2}}}}}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{\dfrac{a^m}{a^n} = a^{m - n}}}}$

$\mathsf{\implies \sqrt[a + b]{(x)^{a^{2} - b^{2}}}\;\times\sqrt[b + c]{x^{b^{2} - c^{2}}}}}\;\times\sqrt[c + a]{x^{c^{2} - a^{2}}}}}}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{\sqrt[n]{a} = a^{\dfrac{1}{n}}}}}$

$\implies \big[(x)^{a^{2} - b^{2}}\big]^{\bigg(\dfrac{1}{a + b}\bigg)}\times\big[(x)^{b^{2} - c^{2}}\big]^{\bigg(\dfrac{1}{b + c}\bigg)}\times\big[(x)^{c^{2} - a^{2}}\big]^{\bigg(\dfrac{1}{c + a}\bigg)}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{\big[a^m\big]^n = a^{mn}}}}$

$\implies \big[x}\big]^{\bigg(\dfrac{a^2 - b^2}{a + b}\bigg)}\times\big[x\big]^{\bigg(\dfrac{b^2 - c^2}{b + c}\bigg)}\times\big[x\big]^{\bigg(\dfrac{c^2 - a^2}{c + a}\bigg)}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{(p + q)(p - q) = p^2 - q^2}}}$

$\implies \big[x}\big]^{\bigg(\dfrac{(a - b)(a + b)}{a + b}\bigg)}\;\times\;\big[x\big]^{\bigg(\dfrac{(b - c)(b + c)}{b + c}\bigg)}\;\times\;\big[x\big]^{\bigg(\dfrac{(c - a)(c + a)}{c + a}\bigg)$

$\implies \big[x}\big]^{a - b}\;\times\;\big[x\big]^{b - c}\;\times\;\big[x\big]^{c - a}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{a^m \times a^n = a^{m + n}}}}$

$\mathsf{\implies \big[x}\big]^{a - b + b - c + c - a}}$

$\mathsf{\implies \big[x\big]^{0}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{a^0 = 1}}}$

$\mathsf{\implies \big[x\big]^{0} = 1}$