Question

Please give me answer of the following sum of Maths​

Answer 1

Answer:

L.H.S= R.H. S (proved)

Answer 2

Question

$\rm \: : \implies \: \bigg( \dfrac{ {x}^{m} }{ {x}^{n} } \bigg) {}^{ \dfrac{1}{mn} } \times \bigg( \dfrac{ {x}^{n} }{ {x}^{r} } \bigg) {}^{ \dfrac{1}{nr} } \times \bigg( \dfrac{ {x}^{r} }{ {x}^{m} } \bigg) {}^{ \dfrac{1}{rm} } = 1$

Solution:-

$\rm \: : \implies \: \bigg( \dfrac{ {x}^{m} }{ {x}^{n} } \bigg) {}^{ \dfrac{1}{mn} } \times \bigg( \dfrac{ {x}^{n} }{ {x}^{r} } \bigg) {}^{ \dfrac{1}{nr} } \times \bigg( \dfrac{ {x}^{r} }{ {x}^{m} } \bigg) {}^{ \dfrac{1}{rm} } = 1$

Using this exponential law

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

$\rm : \implies \: ({x}^{a} ) {}^{b} = {x}^{ab}$

$\rm : \implies \: x {}^{a} \times {x}^{b} = {x}^{a + b}$

$\rm \: : \implies \: {x}^{0} = 1$

Now we get

$\rm : \implies( {x}^{m - n} ) {}^{ \frac{1}{mn} } \times ( {x}^{n- r} ) {}^{ \frac{1}{nr} } \times ( {x}^{r- m} ) {}^{ \frac{1}{rm} }$

$\rm \: : \implies(x) {}^{ \frac{m - n}{mn} } \times (x) {}^{ \frac{n - r}{nr} } \times (x) {}^{ \frac{r - m}{rm} }$

$\rm : \implies(x) {}^{ \frac{m}{mn} - \frac{n}{mn} } \times (x) {}^{ \frac{n}{nr} - \frac{r}{nr} } \times (x) {}^{ \frac{r}{rm} - \frac{m}{rm} }$

$\rm : \implies \: (x) {}^{ \frac{1}{n} - \frac{1}{m} } \times (x) {}^{ \frac{1}{r} - \frac{1}{n} } \times (x) {}^{ \frac{1}{m} - \frac{1}{r} }$

$\rm: \implies(x) {}^{ \frac{1}{n} - \frac{1}{m} + \frac{1}{r} - \frac{1}{n} + \frac{1}{m} - \frac{1}{r} }$

$\rm : \implies \: ( {x})^{0} = 1$

Hence proved