Question

Если x+y+z=1 и 1/x+1/y+1/z=0 , найдите x^2+y^2+z^2 .

Answer 1

Приводим 2-е выражение  к общему знаменателю :

(zy+xz+xy)/xyz=0

x*y*z≠0

xy+yz+xz=0

(x+y+z)^2=x^2+y^2+z^2 +2xy +2yz+2xz=1

x^2+y^2+z^2=1

Answer 2

To find the value of x^2 + y^2 + z^2, we can use some algebraic manipulations based on the given equations:

Given:

1. x + y + z = 1
2. 1/x + 1/y + 1/z = 0

Let's square equation 1 to get x^2 + y^2 + z^2 + 2(xy + yz + zx) = 1.

Now, we need to find the value of xy + yz + zx. To do that, we can square equation 2:

(1/x + 1/y + 1/z)^2 = 0 (1/x^2 + 1/y^2 + 1/z^2) + 2(1/(xy) + 1/(yz) + 1/(zx)) = 0

Now, let's find the value of 1/(xy) + 1/(yz) + 1/(zx). We can use equation 2 and rewrite it as:

1/(xy) + 1/(yz) + 1/(zx) = -1/(x+y+z)^2 1/(xy) + 1/(yz) + 1/(zx) = -1/1^2 1/(xy) + 1/(yz) + 1/(zx) = -1

Now, going back to the squared equation 1:

x^2 + y^2 + z^2 + 2(xy + yz + zx) = 1

We know that xy + yz + zx = -1, so we can substitute this into the equation:

x^2 + y^2 + z^2 + 2(-1) = 1 x^2 + y^2 + z^2 - 2 = 1

Finally, rearrange the equation to solve for x^2 + y^2 + z^2:

x^2 + y^2 + z^2 = 1 + 2 x^2 + y^2 + z^2 = 3

So, the value of x^2 + y^2 + z^2 is 3.