Question

Answer this please. ASAP!

Answer 1

Answer:

hey answer mi please

it's very urgent

Answer 2

Answer:

$\large\underline{\sf{Given- }}$

$\qquad\leadsto\quad \pmb {\mathfrak{\:a: b \: = \: b: c}}$

$\large\underline{\sf{To\:prove - }}$

$\rm :\longmapsto\:\dfrac{a}{a + b} = \dfrac{a - b}{a - c}$

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:a: b \: = \: b: c$

$\rm :\longmapsto\:\dfrac{a}{b} = \dfrac{b}{c} = k$

$\rm :\implies\:b = ck$

and

$\rm :\longmapsto\:a = bk$

$\bf\implies \:a = ck.k = {ck}^{2}$

Consider LHS,

$\rm :\longmapsto\:\dfrac{a}{a + b}$

$\rm \: = \: \: \dfrac{bk}{bk + b}$

$\rm \: = \: \: \dfrac{bk}{b(k + 1)}$

$\rm \: = \: \: \dfrac{k}{k + 1}$

$\bf\implies \:\: \dfrac{a}{a + b} = \: \: \dfrac{k}{ k + 1} - - - (1)$

Now, Consider RHS,

$\rm :\longmapsto\:\dfrac{a - b}{a - c}$

$\rm \: = \: \: \dfrac{c {k}^{2} - ck }{ {ck}^{2} - c}$

$\rm \: = \: \: \dfrac{ck(k - 1)}{ c( {k}^{2} - 1)}$

$\rm \: = \: \: \dfrac{k(k - 1)}{ ( {k}^{2} - 1)}$

$\rm \: = \: \: \dfrac{k(k - 1)}{ (k - 1)(k + 1)}$

$\rm \: = \: \: \dfrac{k}{ k + 1}$

$\bf\implies \:\: \dfrac{a - b}{a - c} = \: \: \dfrac{k}{ k + 1} - - - (2)$

From equation (1) and (2), we concluded that

$\bf :\longmapsto\:\dfrac{a}{a + b} = \dfrac{a - b}{a - c}$

Hence, Proved

Additional Information :-

$\rm :\longmapsto\:\dfrac{a}{b} = \dfrac{c}{d} \: \: \: then$

$\rm :\longmapsto\:\dfrac{a}{c} = \dfrac{b}{d} \: \:is \: called \: alternendo$

$\rm :\longmapsto\:\dfrac{b}{a} = \dfrac{d}{c} \: \:is \: called \: invertendo$

$\rm :\longmapsto\:\dfrac{a + b}{b} = \dfrac{c + d}{d} \: \: is \: called \: componendo$

$\rm :\longmapsto\:\dfrac{a - b}{b} = \dfrac{c - d}{d} \: \: is \: called \: dividendo$