Cho \(\dfrac{{{x^2}}}{{y + z}} + \dfrac{{{y^2}}}{{x + z}} + \dfrac{{{z^2}}}{{x + y}} = 0\) và \(x + y + z \ne 0.\) Tính giá trị của biểu thức \(A = \dfrac{x}{{y + z}} + \dfrac{y}{{x + z}} + \dfrac{z}{{x + y}}\)
Vì \(\dfrac{{{x^2}}}{{y + z}} + \dfrac{{{y^2}}}{{x + z}} + \dfrac{{{z^2}}}{{x + y}} = 0\) nên ta có
\(x + y + z = x + y + z + 0\) \( = x + y + z + \dfrac{{{x^2}}}{{y + z}} + \dfrac{{{y^2}}}{{x + z}} + \dfrac{{{z^2}}}{{x + y}}\)
\( = \left( {x + \dfrac{{{x^2}}}{{y + z}}} \right) + \left( {y + \dfrac{{{y^2}}}{{x + z}}} \right) + \left( {z + \dfrac{{{z^2}}}{{x + y}}} \right)\)
\( = x\left( {1 + \dfrac{x}{{y + z}}} \right) + y\left( {1 + \dfrac{y}{{x + z}}} \right) + z\left( {1 + \dfrac{z}{{x + y}}} \right)\)
\( = x\left( {\dfrac{{x + y + z}}{{y + z}}} \right) + y\left( {\dfrac{{x + y + z}}{{x + z}}} \right) + z\left( {\dfrac{{x + y + z}}{{x + y}}} \right)\)
\( = \left( {x + y + z} \right)\left( {\dfrac{x}{{y + z}} + \dfrac{y}{{x + z}} + \dfrac{z}{{x + y}}} \right)\)
Suy ra \(x + y + z = \left( {x + y + z} \right)\left( {\dfrac{x}{{y + z}} + \dfrac{y}{{x + z}} + \dfrac{z}{{x + y}}} \right)\)
Mà \(x + y + z \ne 0\) nên \(\dfrac{x}{{y + z}} + \dfrac{y}{{x + z}} + \dfrac{z}{{x + y}} = 1\)
Hay \(A = 1.\)
