Cho x + y + z = 1, 1/x + 1/y + 1/z = 0. tính q = x^2 + y^2 + z^2

To solve this problem, we can use the given equations to find the values of x, y, and z, and then calculate q = x^2 + y^2 + z^2.

From the first equation, we have x + y + z = 1.

From the second equation, we have 1/x + 1/y + 1/z = 0.

To simplify the second equation, we can multiply both sides by xyz to get:
yz + xz + xy = 0.

Now, let's square the first equation to get:
(x + y + z)^2 = 1^2
x^2 + y^2 + z^2 + 2xy + 2xz + 2yz = 1.

We can subtract 2(xy + xz + yz) from both sides to get:
x^2 + y^2 + z^2 = 1 - 2(xy + xz + yz).

Now, we need to find the value of xy + xz + yz.

We can square the equation yz + xz + xy = 0 to get:
y^2z^2 + x^2z^2 + x^2y^2 + 2xyz(x + y + z) = 0.

Since we know that x + y + z = 1, we can simplify this equation to:
y^2z^2 + x^2z^2 + x^2y^2 + 2xyz = 0.

Now, let's square the first equation (x + y + z = 1) again to get:
(x + y + z)^2 = 1^2
x^2 + y^2 + z^2 + 2xy + 2xz + 2yz = 1.

We can subtract 2xyz from both sides to get:
x^2 + y^2 + z^2 - 2xyz = 1 - 2xyz.

Now, we can substitute this expression into the previous equation to get:
y^2z^2 + x^2z^2 + x^2y^2 + 2xyz = x^2 + y^2 + z^2 - 2xyz.

Rearranging this equation, we get:
x^2 + y^2 + z^2 = 2xyz - x^2y^2 - x^2z^2 - y^2z^2.

Now, we can substitute this expression into the equation we found earlier for x^2 + y^2 + z^2:
x^2 + y^2 + z^2 = 1 - 2(xy + xz + yz)
= 1 - 2(2xyz - x^2y^2 - x^2z^2 - y^2z^2)
= 1 - 4xyz + 2x^2y^2 + 2x^2z^2 + 2y^2z^2.

Finally, we can substitute the value of xyz from the second equation (1/x + 1/y + 1/z = 0):
xyz = -1.

Substituting this value into the equation for x^2 + y^2 + z^2, we get:
x^2 + y^2 + z^2 = 1 - 4(-1) + 2x^2y^2 + 2x^2z^2 + 2y^2z^2
= 1 + 4 + 2x^2y^2 + 2x^2z^2 + 2y^2z^2
= 5 + 2x^2y^2 + 2x^2z^2 + 2y^2z^2.

Therefore, q = x^2 + y^2 + z^2 = 5 + 2x^2y^2 + 2x^2z^2 + 2y^2z^2.

Note: The value of q depends on the values of x, y, and z, which are not specified in the question.