Cho a + b + c = 0 ; x + y + z = 0 và ax + by + cz = 0. Chứng minh rằng: ax^2 + by^2 + cz^2 = 0.

Hướng dẫn giải

Do \(x + y + z = 0\) nên \(\left\{ \begin{array}{l}x = - \left( {y + z} \right)\\y = - \left( {x + z} \right)\\z = - \left( {x + y} \right)\end{array} \right.\), suy ra \(\left\{ \begin{array}{l}{x^2} = {\left( {y + z} \right)^2}\\{y^2} = {\left( {x + z} \right)^2}\\{z^2} = {\left( {x + y} \right)^2}\end{array} \right..\) 

Do \(a + b + c = 0\) nên \(\left\{ \begin{array}{l}b + c = - a\\c + a = - b\\a + b = - c\end{array} \right..\)

Do \(\frac{a}{x} + \frac{b}{y} + \frac{c}{z} = 0\) nên \(\frac{{ayz + bxz + cxy}}{{xyz}} = 0\) suy ra \(bxz + cxy + ayz = 0\).

Ta có: \(a{x^2} + b{y^2} + c{z^2} = a{\left( {y + z} \right)^2} + b{\left( {x + z} \right)^2} + c{\left( {x + y} \right)^2}\)

\( = a{y^2} + 2ayz + a{z^2} + b{x^2} + 2bxz + b{z^2} + c{x^2} + 2cxy + c{y^2}\)

\( = \left( {b + c} \right){x^2} + \left( {a + c} \right){y^2} + \left( {a + b} \right){z^2} + 2\left( {bxz + cxy + ayz} \right)\)

\( = \left( {b + c} \right){x^2} + \left( {a + c} \right){y^2} + \left( {a + b} \right){z^2}\)

\( = \left( { - a} \right){x^2} + \left( { - b} \right){y^2} + \left( { - c} \right){z^2}\)

Do đó \(2\left( {a{x^2} + b{y^2} + c{z^2}} \right) = 0\) nên \(a{x^2} + b{y^2} + c{z^2} = 0.\)

Vậy \(a{x^2} + b{y^2} + c{z^2} = 0.\)