Question

$\huge \tt \pink{Question: - }$
$यदि \: \frac{x}{(b - c)(b + c - 2a)} = \frac{y}{(c - a)(c + a - 2a)} = \frac{z}{(a - b)(a + b - 2c)} \: है,$
तो x+y+z का मूल्य कया है।

Please help me..​

Answer 1

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

$\sf \: \dfrac{x}{(b - c)(b + c - 2a)} = \dfrac{y}{(c - a)(c + a - 2b)} = \dfrac{z}{(a - b)(a + b - 2c)}$

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

$\rm :\longmapsto\:x + y + z$

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

$\sf \:Let \: \dfrac{x}{(b - c)(b + c - 2a)} = \dfrac{y}{(c - a)(c + a - 2b)} = \dfrac{z}{(a - b)(a + b - 2c)} = k$

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

$\rm :\longmapsto\:x = k(b - c)(b + c - 2a)$

$\rm :\longmapsto\:x = k({b}^{2} - {c}^{2} - 2a(b - c))$

$\rm :\longmapsto\: x=k({b}^{2} - {c}^{2} + 2ac - 2ab) - - - (1)$

Also,

$\rm :\longmapsto\:\dfrac{y}{(c - a)(c + a - 2b)} = k$

$\rm :\longmapsto\:y = k(c - a)(c + a - 2b)$

$\rm :\longmapsto\:y = k( {c}^{2}-{a}^{2}-2bc+2ba) - - (2)$

Again,

$\rm :\longmapsto\:\dfrac{z}{(a - b)(a + b - 2c)} = k$

$\rm :\longmapsto\:z = k(a - b)(a + b - 2c)$

$\rm :\longmapsto\:z = k( {a}^{2} - {b}^{2} - 2ca + 2bc) - - (3)$

Now,

Adding equation (2), (3) and (4), we get

$\rm :\longmapsto\:\: x + y + z$

$\rm \: = k({b}^{2}-{c}^{2}+2ac-2ab+{c}^{2}-{a}^{2}-2bc+2ba+{a}^{2}-{b}^{2}-2ca +2bc)$

$\rm \: = \: k \: \times \: 0$

$\rm \: = \: 0$

Answer 2

Answer:

okk dear ...the answer of this question will be like this......