Question

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Steps needed..

If |a-b| =1 ; |b-c| =1 ; |c-a| = 2 and abc= 60
Find a/bc + b/ca + c/ab - 1/a - 1/b - 1/c​

Answer 1

Given

|a - b| = 1

|b - c| = 1

|c - a| = 1

abc = 60

To find

a/bc + b/ca + c/ab - 1/a - 1/b - 1/c

Solution

To remove modulus, square both sides

|a - b|² = 1²

→ a² + b² - 2ab = 1

|b - c|² = 1²

→ b² + c² - 2bc = 1

|c - a|² = 2²

→ c² + a² - 2ac = 4

Now, divide the three equations by abc

a²/abc + b²/abc - 2ab/abc = 1/abc

→ a/bc + b/ac - 2/c = 1/60 (since abc = 60)

Similarly, you'll get from the other two equations,

b/ac + c/ab - 2/a = 1/60

and

c/ab + a/bc - 2/b = 4/60

Add the three equations, we get

2(a/bc + b/ac + c/ab - 1/c - 1/a - 1/b) = 6/60

→ a/bc + b/ac + c/ab - 1/c - 1/a - 1/b = 1/20

Hence 1/20 is the answer. :)

Answer 2

Step-by-step explanation:

abc- 69

The current passing through it is (a) 2A (b) 4A (c) 8A (d) 16A. Answer.

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