Cho a, b, c là các số hữu tỉ khác 0 tỏa mãn a + b + c = 0. Chứng minh rằng: 1/a^2 + 1/b^2

\[\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} = {\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right)^2} - 2.\left( {\frac{1}{{ab}} + \frac{1}{{bc}} + \frac{1}{{ca}}} \right)\]

\[ = {\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right)^2} - 2.\frac{{a + b + c}}{{abc}} = {\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right)^2} - 2.\frac{0}{{abc}}\]

\[ = {\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right)^2}\] (đpcm)