Question

If

$\frac{a}{b} = \frac{c}{d} = \frac{e}{f}$

Then prove that,

$\frac{2 {a}^{4} {b}^{2} + 3 {a}^{2} {e}^{2} - 5 {e}^{4} f }{2 {b}^{6} + 3 {b}^{2} {f}^{2} - 5 {f}^{5} } = \frac{ {a}^{4} }{ {b}^{4} }$

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Answer 1

Answer:

Step-by-step explanation:

Holla user

Hope thse helps u to understand my ans ^_^

At last LHS = RHS

Hence proved ^_^

Answer 2

Answer:

Given :

$\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f}$

Let a/b = r .

Then a = b r .

c = d r .

e = f r .

$\dfrac{2 {a}^{4} {b}^{2}+ 3 {a}^{2} {e}^{2} - 5 {e}^{4}f}{2 {b}^{6} + 3 {b}^{2} {f}^{2} - 5 {f}^{5}} =\dfrac{{a}^{4} }{ {b}^{4} } \\\\\implies \dfrac{2r^4b^6+3r^4b^2d^4-5r^4f^5}{2b^6+3b^2d^2-5f^5}\\\\\implies \dfrac{r^4(2b^6+3b^2d^2-5f^5)}{2b^6+3b^2d^2-5f^5}\\\\\implies r^4\\\\\implies \dfrac{a^4}{b^4}$

Step-by-step explanation:

Firstly take all the values and substitute all fractions with r.

Then substitute in the LHS the value of a , c and e to get the results .

At last take common and cancel .

Use the laws of indices : $a^b\times a^c=a^{b+c}$ .

Remember another thing ,

Using the laws of ratio :

Each ratio = sum of antecedents / sum of consequents .