Math, asked by salmansk01, 1 year ago

if a+b+c=0 then prove that a2/(2a2+bc)+b2/(2b2+ac)+c2/(2c2+ab)=1

Answers

Answered by spn62120
2
You can write your expression as: 
1/(2a2+bc/a2)+1/(2+ca/b2)+1/(2+ab/c2)=... which reduces to: 
a2/bc + b2/ac + c2/ab = 3. If you now plug a = -(b+c) you will get: 
(b+c)^2/bc - (b^2/c + c^2/b)/(b+c) = 3, which reduces to 
[(b+c)^3 - b^3 - c^3]/[bc*(b+c)] = 3*bc*(b+c)/[bc*(b+c)] which is 
indeed equal 3 as needed.
Answered by lakshaysethi2006
2

Answer:

a+c+b  = 0

a = -b-c

ca = -bc - c²

2b² + ca = b² - c² + b² - bc

= - (b-c)(a-b)

lly, 2a² + bc = -(a-b)(c-a) and 2c² + ab = -(c-a)(b-c)

Now LHS = -a²/(a-b)(c-a) - b²/(b-c)(a-b) - c²/ (c-a) (b-c)

now by LCM and multiplication

= -{a²b-a²c+b²c-ab²+ac²-bc² / - (a²b-a²c+b²c -ab² + ac² - bc²)]

= 1 = RHS 

Hence Proved.

Read more on Brainly.in - https://brainly.in/question/5217#readmore

Step-by-step explanation:

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