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Inessa [10]
3 years ago
15

Show that:

%7D%20%7D%20.%20%5Csqrt%5Bbc%5D%7B%20%5Cfrac%7B%20%7Bx%7D%5E%7Bb%7D%20%7D%7B%20%7Bx%7D%5E%7Bc%7D%20%7D%20%7D%20.%20%5Csqrt%5Bca%5D%7B%20%5Cfrac%7B%20%7Bx%7D%5E%7Bc%7D%20%7D%7B%20%7Bx%7D%5E%7Ba%7D%20%7D%20%7D%20%20%20%3D%201." id="TexFormula1" title="\sqrt[ab]{ \frac{ {x}^{a} }{ {x}^{b} } } . \sqrt[bc]{ \frac{ {x}^{b} }{ {x}^{c} } } . \sqrt[ca]{ \frac{ {x}^{c} }{ {x}^{a} } } = 1." alt="\sqrt[ab]{ \frac{ {x}^{a} }{ {x}^{b} } } . \sqrt[bc]{ \frac{ {x}^{b} }{ {x}^{c} } } . \sqrt[ca]{ \frac{ {x}^{c} }{ {x}^{a} } } = 1." align="absmiddle" class="latex-formula">
Brainliest! Brainliest..ans.
​
Mathematics
2 answers:
nika2105 [10]3 years ago
7 0

Here we need to Prove that ,

\bf\implies \sqrt[ab]{ \frac{ {x}^{a} }{ {x}^{b} } } . \sqrt[bc]{ \frac{ {x}^{b} }{ {x}^{c} } } . \sqrt[ca]{ \frac{ {x}^{c} }{ {x}^{a} } } = 1.

Lets start with LHS , which is ,

\bf\implies LHS = \sqrt[ab]{ \dfrac{ {x}^{a} }{ {x}^{b} } } . \sqrt[bc]{ \dfrac{ {x}^{b} }{ {x}^{c} } } . \sqrt[ca]{ \dfrac{ {x}^{c} }{ {x}^{a} } }

We know that ,

\bf\implies \sqrt[n]{k^m }= k^{\frac{m}{n}}

• So that ,

\bf\implies \sqrt[ab]{x^{a-b}} . \sqrt[bc]{x^{b-c}} .\sqrt[ac]{x^{c-a}} \\\\\bf\implies x^{\frac{a-b}{ab}} . x ^{\frac{b-c}{bc}} .x^{\frac{c-a}{ca}} \\\\\bf\implies x^{\frac{ ac - bc + ab - ac + bc - ba }{abc } }\\\\\bf\implies x^{\frac{0}{abc}} \\\\\bf\implies x^0 \\\\\bf\implies 1 = RHS

Hence ,

\implies \boxed{\bf \red{ LHS = RHS }}

victus00 [196]3 years ago
7 0

\large\underline{\sf{Solution-}}

Consider <u>LHS</u>:

\sf= \sqrt[\sf ab]{\dfrac{\sf x^{a}}{\sf x^{b}}}\cdot \sqrt[\sf bc]{\dfrac{\sf x^{b}}{\sf x^{c}}}\cdot \sqrt[\sf ac]{\dfrac{\sf x^{c}}{\sf x^{a}}}

<em>As we know that:</em>

\sf\red\leadsto\dfrac{x^{a}}{x^{b}}=x^{a-b}

<u>We get:</u>

\sf= \sqrt[\sf ab]{\sf x^{a-b}}\cdot \sqrt[\sf bc]{\sf x^{b-c}}\cdot \sqrt[\sf ac]{\sf x^{c-a}}

<u>We can also write it as:</u>

\sf= x^{\dfrac{a-b}{ab}}\cdot x^{\dfrac{b-c}{bc}}\cdot x^{\dfrac{c-a}{ac}}

<em>We know that:</em>

\sf \red\leadsto x^{a}\cdot x^{b}=x^{a+b}

<u>We get:</u>

\sf= x^{\bigg(\dfrac{a-b}{ab}+\dfrac{b-c}{bc}+\dfrac{c-a}{ac}\bigg)}

\sf= x^{\bigg(\dfrac{c(a-b)+a(b-c)+b(c-a)}{abc}\bigg)}

\sf= x^{\bigg(\dfrac{ac-bc+ab-ac+bc-ab}{abc}\bigg)}

\sf= x^{\bigg(\dfrac{0}{abc}\bigg)}

\sf= x^{0}

\sf=1

\textsf{= RHS}\\

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