Question

show that : follow the image.
Pls help ​

Answer 1

$\sqrt{ {x}^{p - q} } \times \sqrt{ {x}^{q - r} } \times \sqrt{ {x}^{r - p} } \\ \\ = \frac{ \sqrt{ {x}^{p} } }{ \sqrt{ {x}^{q} } } \times \frac{ \sqrt{ {x}^{q} } }{ \sqrt{ {x}^{r} } } \times \frac{ \sqrt{ {x}^{r} } }{ \sqrt{ {x}^{p} } } \\ \\ = 1 \times 1 \times 1 \\ \\ = 1$

Answer 2

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{\bf{Question}}}}}}}}$

$\longrightarrow \sf proof :- \sqrt{x^{p-q}} \times \sqrt{x^{q-r}}\times\sqrt{x^{r-p}} = 1$

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{\bf{Answer}}}}}}}}$

LHS:-

$\longrightarrow \sf \sqrt{x^{p-q}} \times \sqrt{x^{q-r}}\times\sqrt{x{r-p}}$

${\bold{\blue{\boxed{\bf{x^a-b = \dfrac{x^a}{x^b}}}}}}$

$\longrightarrow\sf \sqrt{\dfrac{x^p}{x^q}}\times\sqrt{\dfrac{x^q}{x^r}}\times\sqrt{\dfrac{x^r}{x^p}}$

${\bold{\blue{\boxed{\bf{x^a\times x^b = x^{a+b}}}}}}$

$\longrightarrow\sf\sqrt\dfrac{x^{p+q+r}}{x^{p+q+r}}$

$\longrightarrow\sf\sqrt{x^{\cancel p+\cancel q+\cancel r-\cancel p-\cancel q\cancel -r}}$

$\longrightarrow\sf\sqrt{x^0}$

${\bold{\blue{\boxed{\bf{x^0 = 1}}}}}$

$\longrightarrow\sf\sqrt{1}$

$\longrightarrow\sf\sqrt{1\times1}$

$\longrightarrow\sf\ 1$

${\bold{\blue{\boxed{\bf{\dag 1}}}}}$

RHS:-

${\bold{\blue{\boxed{\bf{\dag 1}}}}}$

COMPARE :-

${\bold{\blue{\boxed{\bf{\dag 1 = 1 }}}}}$

${\bold{\blue{\boxed{\bf{\dag LHS = RHS}}}}}$

${\bold{\blue{\boxed{\bf{\dag Hence\:proved}}}}}$

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