Question

if a+b+c=0 prove that (bc/2a^2+bc)+(ca/2b^2+ca)+(ab/2c^2+ab)=1​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

$\large{ \green{ \sf \: ✿ \: GIVEN \: \: QUESTION : - }}$

$\sf \: if \: (a + b + c) = 0$

$\large{ \green{ \sf \: ✿ \: SHOW \: \: THAT : - }}$

$\sf \bigg(\frac{bc}{2 {a}^{2} + bc}\bigg) + \bigg( \frac{ca}{2 {b}^{2} + ca} \bigg) + \bigg( \frac{ab}{2 {c}^{2} + ab} \bigg) = 1$

$\large{ \green{ \sf \: ✿ \:TAKING \: \: \: LHS. : - }}$

$\sf \bigg(\frac{bc}{2 {a}^{2} + bc}\bigg) + \bigg( \frac{ca}{2 {b}^{2} + ca} \bigg) + \bigg( \frac{ab}{2 {c}^{2} + ab} \bigg)$

$\sf \: we \: have \: the \: relation -$

$\sf \: (a + b + c) = 0$

$\sf \: Taking \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf \: a = - (b + c) \:$

$\sf \: Similarly \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf \: b = - (a + c) \: \: \: and \\ \sf \: c = - (a + b) \: \: \: \: \: \: \: \: \: \:$

$\sf \: Now \: \: from \: \: the \: \: LHS$

$\sf \: 2{a}^{2} + bc = {a}^{2} + {a}^{2} + bc$

$\: \sf \: = {a}^{2} + a( - (b + c)) + bc \\ = \sf {a}^{2} - ab - ac + bc \: \: \: \: \: \: \: \\ \sf \: = a(a - b) - c(a - b) \: \: \: \: \: \\ = \sf (a - b)(a - c) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\sf \: SIMILARLY$

$\sf \: 2 {b}^{2} + ca = (b - c)(b - a)$

$\sf \: 2 {c}^{2} + ab = (c - a)(c - b)$

$\sf \: Putting \: \: in \: \: LHS.$

$\sf \bigg(\frac{bc}{(a - b)(a - c)}\bigg) + \bigg( \frac{ca}{(b - c)(b - a)} \bigg) \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: + \bigg( \frac{ab}{(c - a)(c - b)} \bigg)$

$\sf \: Rearrange \: \: this$

$\sf \frac{bc}{(a - b)(a - c)} - \frac{ca}{( b - c)(a - b)} \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: + \frac{ab}{(a - c)(b - c)}$

$\sf \: = \frac{bc(b - c) - ca(a - b) + ab (c - a)}{(a - b)(b - c)(c - a)}$

$\sf \: = \frac{( {b}^{2}c - {bc}^{2} - c {a}^{2} + abc + abc - {a}^{2}b )}{(a - b)(b - c)(c - a)}$

$\sf \: = \frac{( c( {b}^{2} - {a}^{2} ) + bc(a - b) + ab(c - a)}{(a - b)(b - c)(c - a)}$

$\sf \: = \frac{( - c( (a + b)(a - b) ) + bc(a - b) + ab(c - a)}{(a - b)(b - c)(c - a)}$

$= \sf \: \frac{(a - b)( - c(a + b) + bc) + ab(c - a)}{(a - b)(b - c)(c - a)}$

$= \sf \: \frac{(a - b)( - ca - c b + bc) + ab(c - a)}{(a - b)(b - c)(c - a)}$

$= \sf \: \frac{ - ca(a - b)+ ab(c - a)}{(a - b)(b - c)(c - a)}$