Question

please can anybody solve this question.It is really urgent please. ​

Answer 1

Solution:

Given to prove:

$\tt\longrightarrow \dfrac{aceg}{bdfh}=\dfrac{a^4+c^4+e^4+g^4}{b^4+d^4+f^4+h^4}$

Where:

$\tt\longrightarrow \dfrac{a}{b}=\dfrac{c}{d}=\dfrac{e}{f}=\dfrac{g}{h}$

And the values b, d, f, h ≠ 0

Let us assume that:

$\tt\longrightarrow \dfrac{a}{b}=\dfrac{c}{d}=\dfrac{e}{f}=\dfrac{g}{h}=k$

Where k is some real number, k ≠ 0

Then we can say:

$\tt\longrightarrow a=bk$

$\tt\longrightarrow c=dk$

$\tt\longrightarrow e=fk$

$\tt\longrightarrow g=hk$

Now consider Left Hand Side, we have:

$\tt=\dfrac{aceg}{bdfh}$

Can be written as:

$\tt=\dfrac{(bk)\cdot(dk)\cdot(fk)\cdot(hk)}{bdfh}$

$\tt=\dfrac{k^4\cdot bdfh}{bdfh}$

$\tt=k^4$

Now consider Right Hand Side, we have:

$\tt=\dfrac{a^4+c^4+e^4+g^4}{b^4+d^4+f^4+h^4}$

$\tt=\dfrac{(bk)^4+(dk)^4+(fk)^4+(hk)^4}{b^4+d^4+f^4+h^4}$

$\tt=\dfrac{k^4(b^4+d^4+f^4+h^4)}{b^4+d^4+f^4+h^4}$

$\tt=k^4$

We observe that LHS = RHS

Therefore:

$\tt\longrightarrow \dfrac{aceg}{bdfh}=\dfrac{a^4+c^4+e^4+g^4}{b^4+d^4+f^4+h^4}$

Hence Proved..!!

Learn More:

If a : b and c : d are two ratios such that a : b : : c : d. Then the following results hold true.

1. Invertendo.

$\tt\longrightarrow b : a :  : d : c$

2. Alternendo.

$\tt\longrightarrow a : c :  : b : d$

3. Componendo.

$\tt\longrightarrow (a + b) : b:  : (c + d) : d$

4. Dividendo.

$\tt\longrightarrow (a  -  b) : b:  : (c - d) : d$

5. Componendo and dividendo.

$\tt\longrightarrow(a + b) : (a - b):  : (c  +  d) :(c - d)$

6. Convertendo.

$\tt\longrightarrow a : (a - b):  :c:(c - d)$