Question

If x^2=y+z, y^2=z+x, z^2=x+y, then 1/(x+1)+1/(y+1)+1/(z+1)=

Answer 1

Consider the given equation:

Multiply both the numerator and denominator of the first term by x, second term by y, third term by z. Doing that we get:

Now consider x2=y+zx2=y+z
Adding x on both sides of the equation, we get:
x+x2x2=x+y+z
Taking the reciprocal and multiplying by x on both sides:
x/(x+x2x2) = x/(x+y+z)
Doing the same for the other terms we get:
y/(x+y+z) and z/(x+y+z)

Adding all three terms we get : (x+y+z)/(x+y+z) = 1
So in the case when x,y,z not equal to 0 we get 1

If we consider the case when one of the terms can be 0, then the answer would be 3

So the answer to this question will be either 1 or 3

Answer 2

Answer:

So the answer to this question will be either 1 o

Step-by-step explanation:

Consider the given equation:

Multiply both the numerator and denominator of the first term by x, second term by y, third term by z. Doing that we get:

Now consider x2=y+zx2=y+z

Adding x on both sides of the equation, we get:

x+x2x2=x+y+z

Taking the reciprocal and multiplying by x on both sides:

x/(x+x2x2) = x/(x+y+z)

Doing the same for the other terms we get:

y/(x+y+z) and z/(x+y+z)

Adding all three terms we get : (x+y+z)/(x+y+z) = 1

So in the case when x,y,z not equal to 0 we get 1

If we consider the case when one of the terms can be 0, then the answer would be 3

So the answer to this question will be either 1 o