Question

step by step plz....​

Answer 1

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Answer 2

Answer:

$\displaystyle{\implies \text{Mean proportional} = \left[\frac{(x+y)}{(xy)}\right]}$

Step-by-step explanation:

Given :

$\displaystyle{\frac{x+y}{x-y} , \frac{x^2-y^2}{x^2y^2}}$

We have to find mean proportional ,

We know mean is equal to  extreme

Let mean be a .

Now ,

$\displaystyle{a\times a=\frac{x+y}{x-y}\times \frac{x^2-y^2}{x^2y^2}}\\\\\\\display \text{Using identity here p^2-q^2=(p+q)(p-q) }\\\\\displaystyle{a^2=\frac{(x+y)}{(x-y)}\times \frac{(x-y)(x+y)}{x^2y^2}}\\\\\\\displaystyle{a^2=\frac{(x+y)^2}{(x^2y^2)}}\\\\\\\displaystyle{a^2=\left[\frac{(x+y)}{(xy)}\right]^2}\\\\\\\displaystyle{a=\left[\frac{(x+y)}{(xy)}\right]^{2\times1/2}}\\\\\\\displaystyle{a=\left[\frac{(x+y)}{(xy)}\right]}$

$\displaystyle{\implies \text{Mean proportional} = \left[\frac{(x+y)}{(xy)}\right]}$

Hence we get answer.