Question

pls answer to my question​

Answer 1

Topic :-

Exponents and Powers

To Prove :-

$\dfrac{x^{m+n} \times x^{n+l} \times x^{l+m}}{(x^m \times x^n \times x^l)^2}=1$

Formula to be Used :-

$x^a \times x^b = x^{a+b}$

$(x^a)^b=x^{ab}$

Solution :-

Solving LHS,

$\dfrac{x^{m+n} \times x^{n+l} \times x^{l+m}}{(x^m \times x^n \times x^l)^2}$

Using Formula,

$\dfrac{x^{m+n} \times x^{n+l} \times x^{l+m}}{(x^m \times x^n \times x^l)^2}$

$\dfrac{x^{m+n+n+l+l+m}}{(x^{m+n+l})^2}$

$\dfrac{x^{2m+2n+2l}}{x^{2(m+n+l)}}$

$\dfrac{x^{2m+2n+2l}}{x^{2m+2n+2l}}$

1

RHS,

1

We observe that LHS = RHS.

Hence, Proved !!

Additional Formulae :-

$\dfrac{x^a}{x^b}=x^{a-b}$

$x^0=1$

$x^a \times y^a = (xy)^a$

Answer 2

Question :-

$\underline{show \: \: that \: :} \\ \\ (i) \: \: \frac{ {x}^{m + n} \times {x}^{n + l} \times {x}^{l + m} }{( {x}^{m} \times {x}^{n} \times {x}^{l} )^{2}} = 1$

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Solution :-

To Find :-

$\dfrac{x^{m+n} \times x^{n+l} \times x^{l+m}}{(x^m \times x^n \times x^l)^2}=1$

Formula Used :-

$\sf{x^a \times x^b = x^{a+b}}$

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

Calculation :-

Firstly, Solving Left Hand Side (LHS),

$\sf\implies\dfrac{x^{m+n} \times x^{n+l} \times x^{l+m}}{(x^m \times x^n \times x^l)^2}$

Using the Formula,

$\sf\implies\dfrac{x^{m+n} \times x^{n+l} \times x^{l+m}}{(x^m \times x^n \times x^l)^2}$

$\sf\implies\dfrac{x^{m+n+n+l+l+m}}{(x^{m+n+l})^2}$

$\sf\implies\dfrac{x^{2m+2n+2l}}{x^{2(m+n+l)}}$

$\sf\implies\dfrac{x^{2m+2n+2l}}{x^{2m+2n+2l}}$

$\bf\implies \red1$

Now, Solving Right Hand Side (RHS),

$\bf\implies \red1$

Observation :-

Since, Left Hand Side (LHS) = Right Hand Side (RHS)

Hence, Proved.

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