Question

(2-¹+3-¹)-²
Sol of this qs​

Answer 1

Step-by-step explanation:

Given Question :

$\sf \dashrightarrow( {2}^{ - 1} + {3}^{ - 1} ) {}^{ - 2}$

Solution :

$\sf \dashrightarrow( {2}^{ - 1} + {3}^{ - 1} ) {}^{ - 2}$

We know that,

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

Using this,

$\sf \dashrightarrow \bigg( \dfrac{1}{ {2}^{1} } + \dfrac{1}{ {3}^{1} } {\bigg)}^{ - 2}$

$\sf \dashrightarrow \bigg( \dfrac{1}{ {2}^{} } + \dfrac{1}{ {3}^{} } {\bigg)}^{ - 2}$

Taking L.C.M.,

$\sf \dashrightarrow \bigg( \dfrac{1 \times 3}{ {2 \times 3}^{} } + \dfrac{1 \times 2}{ {3 \times 2}^{} } {\bigg)}^{ - 2}$

$\sf \dashrightarrow \bigg( \dfrac{3}{ {6}^{} } + \dfrac{2}{ {6}^{} } {\bigg)}^{ - 2}$

$\sf \dashrightarrow \bigg( \dfrac{3 + 2}{ {6}^{} } {\bigg)}^{ - 2}$

$\sf \dashrightarrow \bigg( \dfrac{5}{ {6}^{} } {\bigg)}^{ - 2}$

We know,

$\bf \dashrightarrow \bigg (\dfrac{a}{b} {\bigg)}^{ - c} = \bigg (\dfrac{b}{a} {\bigg)}^{ c}$

Using this,

$\sf \dashrightarrow \bigg( \dfrac{6}{ {5}^{} } {\bigg)}^{ 2}$

$\sf \dashrightarrow \dfrac{6}{ {5}^{} } \times \dfrac{6}{5}$

$\blue{ \sf \dashrightarrow \boxed{ \bf\dfrac{36}{ {25}^{} } } \: \bigstar}$