Đáp án:
$S = -\dfrac32$
Giải thích các bước giải:
Ta có:
$\quad x + y + z = 0$
$\to \begin{cases}y + z = - x\\z + x = -y\\x + y = - z\end{cases}$
$\to \begin{cases}(y+z)^2 = x^2\\(z+x)^2 = y^2\\(x+y)^2 = z^2\end{cases}$
$\to \begin{cases}y^2 + z^2 = x^2 - 2yz\\z^2 + x^2 = y^2 - 2zx\\x^2 + y^2 = z^2 - 2xy\end{cases}$
Ta được:
$S =\dfrac{x^2}{y^2 + z^2 - x^2} +\dfrac{y^2}{z^2 + x^2 - y^2} +\dfrac{z^2}{x^2 + y^2 - z^2}$
$\to S = \dfrac{x^2}{x^2 - 2yz - x^2} +\dfrac{y^2}{y^2 - 2zx - y^2} +\dfrac{z^2}{z^2 - 2xy - z^2}$
$\to S =- \dfrac{x^2}{2yz} -\dfrac{y^2}{2zx} -\dfrac{z^2}{xy}$
$\to S = -\dfrac12\left(\dfrac{x^2}{yz} +\dfrac{y^2}{zx} +\dfrac{z^2}{xy}\right)$
$\to S = -\dfrac12\cdot \dfrac{x^3 + y^3 + z^3}{xyz}$
$\to S = -\dfrac12\cdot\dfrac{x^3 + y^3 + z^3 - 3xyz + 3xyz}{xyz}$
$\to S = -\dfrac12\cdot\left[\dfrac{(x+y+z)(x^2 + y^2 + z^2 - xy - yz - zx)}{xyz} + 3\right]$
$\to S = -\dfrac12\cdot (0 + 3)$
$\to S = -\dfrac32$
