[tex]\huge \tt \pink{Question: – } \\ [/tex]
[tex]यदि \: \frac{x}{(b – c)(b + c – 2a)} = \frac{y}{(c – a)(c + a – 2a

2 thoughts on “<br />[tex]\huge \tt \pink{Question: – } \\ [/tex]<br />[tex]यदि \: \frac{x}{(b – c)(b + c – 2a)} = \frac{y}{(c – a)(c + a – 2a”

1. Answer:

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

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2. [tex]\large\underline{\sf{Given- }}[/tex]

   [tex] \sf \: \dfrac{x}{(b – c)(b + c – 2a)} = \dfrac{y}{(c – a)(c + a – 2b)} = \dfrac{z}{(a – b)(a + b – 2c)} [/tex]

   [tex]\large\underline{\sf{To\:Find – }}[/tex]

   [tex]\rm :\longmapsto\:x + y + z[/tex]

   [tex]\large\underline{\sf{Solution-}}[/tex]

   [tex] \sf \:Let \: \dfrac{x}{(b – c)(b + c – 2a)} = \dfrac{y}{(c – a)(c + a – 2b)} = \dfrac{z}{(a – b)(a + b – 2c)} = k[/tex]

   [tex]\rm :\implies\:\dfrac{x}{(b – c)(b + c – 2a)} = k[/tex]

   [tex]\rm :\longmapsto\:x = k(b – c)(b + c – 2a)[/tex]

   [tex]\rm :\longmapsto\:x = k({b}^{2} – {c}^{2} – 2a(b – c)) [/tex]

   [tex]\rm :\longmapsto\: x=k({b}^{2} – {c}^{2} + 2ac – 2ab) – – – (1)[/tex]

   Also,

   [tex]\rm :\longmapsto\:\dfrac{y}{(c – a)(c + a – 2b)} = k[/tex]

   [tex]\rm :\longmapsto\:y = k(c – a)(c + a – 2b)[/tex]

   [tex]\rm :\longmapsto\:y = k( {c}^{2}-{a}^{2}-2bc+2ba) – – (2)[/tex]

   Again,

   [tex]\rm :\longmapsto\:\dfrac{z}{(a – b)(a + b – 2c)} = k[/tex]

   [tex]\rm :\longmapsto\:z = k(a – b)(a + b – 2c)[/tex]

   [tex]\rm :\longmapsto\:z = k( {a}^{2} – {b}^{2} – 2ca + 2bc) – – (3) [/tex]

   Now,

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

   [tex] \rm :\longmapsto\:\: x + y + z [/tex]

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

   [tex] \rm \: = \: k \: \times \: 0[/tex]

   [tex] \rm \: = \: 0[/tex]

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