Question

please tell the answer​

Answer 1

Question :-

The Value of $\Big [\Big (\dfrac{1}{4}\Big )^{-2} + \Big (\dfrac{1}{3}\Big )^{-2} \Big ] \div \Big (\dfrac{1}{5}\Big )^{-2}$is

$\begin {gathered} \end {gathered}$

Solution :-

$\begin {aligned} & \bold {\Big [\Big (\dfrac{1}{4}\Big )^{-2} + \Big (\dfrac{1}{3}\Big )^{-2} \Big ] \div \Big (\dfrac{1}{5}\Big )^{-2}} \Rightarrow \Big [\Big (\dfrac{4}{1}\Big )^{2} + \Big (\dfrac{3}{1}\Big )^{2} \Big ] \div \Big (\dfrac{5}{1}\Big )^{2}\\\\\\& \Rightarrow \Big [(4)^{2} + (3)^{2} \Big ] \div (5)^{2} = \dfrac{(4)^{2} + (3)^{2}}{(5)^{2}}\\\\\\& \Rightarrow \dfrac{(16 + 9)}{25} \Rightarrow \dfrac{25}{25} = \bold 1 \end {aligned}$

$\begin {gathered} \end {gathered}$

Laws Used :-

i. $a^{-m} = \dfrac{1}{a^m}$

$\begin {gathered} \end {gathered}$

Additional Information :-

Some more Laws of Exponents :

i. $a^m \times a^n = a^{m + n}$

ii. $\dfrac{a^m}{a^n} = a^{m - n}$

iii. $(a^m)^n = a^{m \times n}$

iv. $a^m \times b^m = (a \times b)^m$

v. $\Big (\dfrac{a}{b}\Big )^m = \dfrac{a^m}{b^m}$

vi. $a^0 = 1$

Answer 2

$\large\underline\purple{\bold{Solution :- }}$

$\tt \: \bigg[ {\bigg( \dfrac{1}{4} \bigg)^{ - 2} \: + \bigg( \dfrac{1}{3} \bigg)}^{ - 2} \bigg] \div \bigg( \dfrac{1}{5} \bigg)^{ - 2}$

$\tt\implies \:\bigg[ {(4)}^{2} + {(3)}^{2} \bigg] \div {(5)}^{2}$

$\tt\implies \: \bigg[ 16 + 9\bigg] \div {(5)}^{2}$

$\tt\implies \:25 \div {(5)}^{2}$

$\tt\implies \:25 \times \dfrac{1}{25}$

$\tt\implies \:1$

$\large \red{\bf \: ⟼ Explore \: more }$

$\begin{gathered}(1)\:{\underline{\boxed{\bf{\blue{a^m\times{a^n}\:=\:a^{m\:+\:n}\:}}}}} \\ \end{gathered}$

$\begin{gathered}(2)\:{\underline{\boxed{\bf{\purple{\dfrac{a^m}{a^n}\:=\:a^{m\:-\:n}\:}}}}} \\ \end{gathered}$

$\begin{gathered}(3)\:{\underline{\boxed{\bf{\orange{\dfrac{1}{x^n}\:=\:x^{-n}\:}}}}} \\ \end{gathered}$

$\begin{gathered}(4)\:{\underline{\boxed{\bf{\color{peru}{(a^m)^n\:=\:a^{m\times{n}}\:}}}}} \\ \end{gathered}$

$\begin{gathered}(5)\:{\underline{\boxed{\bf{\red{ {x}^{0} = 1}}}}} \\ \end{gathered}$