If a(x+y) = b(x-y) = 2ab,then the value of 2( | vedclass

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(D) Given: $a(x+y) = 2ab$ and $b(x-y) = 2ab$.From $a(x+y) = 2ab$,we get $x+y = 2b$  $(i)$.From $b(x-y) = 2ab$,we get $x-y = 2a$  $(ii)$.Adding $(i)$ a

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$4(a^2 + b^2)$

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(D) Given: $a(x+y) = 2ab$ and $b(x-y) = 2ab$.From $a(x+y) = 2ab$,we get $x+y = 2b$  $(i)$.From $b(x-y) = 2ab$,we get $x-y = 2a$  $(ii)$.Adding $(i)$ and $(ii)$:$(x+y) + (x-y) = 2b + 2a$$2x = 2(a+b) \implies x = a+b$.Subtracting $(ii)$ from $(i)$:$(x+y) - (x-y) = 2b - 2a$$2y = 2(b-a) \implies y = b-a$.Now,calculate $2(x^2 + y^2)$:$2((a+b)^2 + (b-a)^2)$$= 2(a^2 + b^2 + 2ab + b^2 + a^2 - 2ab)$$= 2(2a^2 + 2b^2)$$= 4(a^2 + b^2)$.

If $a(x+y) = b(x-y) = 2ab$,then the value of $2(x^2 + y^2)$ is:

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