+) \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)
\(=\frac{c+a+b}{abc}=\frac{0}{abc}=0\)
+) \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
\(=\frac{\left(bc+ac+ab\right)^2}{\left(abc\right)^2}=\frac{b^2c^2+2bc\left(ac+ab\right)+\left(ac+ab\right)^2}{a^2b^2c^2}\)
\(=\frac{b^2c^2+2abc^2+2ab^2c+a^2c^2+2a^2bc+a^2b^2}{a^2b^2c^2}\)
\(=\frac{b^2c^2+a^2c^2+a^2b^2+2abc\left(a+b+c\right)}{a^2b^2c^2}\)
\(=\frac{b^2c^2+a^2c^2+a^2b^2}{a^2b^2c^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
a/ Với a, b, c khác 0 ta có:
\(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{c}{abc}+\frac{a}{abc}+\frac{b}{abc}=\)\(\frac{a+b+c}{abc}=\frac{0}{abc}=0\) (dpcm)
