Question

(b+c)/(a-b)(a-c)+(c+a)/(b-c)(b-a)+(a+b)/(c-a)(c-b).​

Answer 1

Answer:

1

Step-by-step explanation:

is the correct answer

hope it helps you

Answer 2

Step-by-step explanation:

(b+c) / (a-b) (a-c) + (c+a) / (b-a) (b-c) + (a+b) /(c-a) (c-b)

= (b+c) /(a-b) (a-c) - (a+c) /(a-b) (b-c) + (a+b) /(a-c) (b-c)

= [(b+c) (b-c) - (a+c) (a-c) + (a+b) (a-b) ] /(a-b) (b-c) (a-c)

= [ (b²-c²) - (a²-c²) + (a²-b²) ] / (a-b) (b-c) (a-c)

=[b²-c²-a²+c²+a²-b²] / (a-b) (b-c) (a-c)

= 0/ (a-b) (b-c) (a-c)

= infinity ..................answer

=> 0/something is always equal to " 1

Hope you are Sastifed