Rút gọn biểu thức sau: B = (ab + bc + ca) (1/a +1/b + 1/c) - abc (1/a^2 + 1/b^2 + 1/c^2)

Hướng dẫn giải

Với \(a,\,\,b,\,\,c \ne 0,\) ta có

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

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

\( = b + a + \frac{{ab}}{c} + \frac{{bc}}{a} + c + b + c + \frac{{ca}}{b} + a - \frac{{bc}}{a} - \frac{{ac}}{b} - \frac{{ab}}{c}\)

\( = 2\left( {a + b + c} \right).\)

Vậy \(B = 2\left( {a + b + c} \right).\)