Cho ba số a,b,c đôi một khác nhau. Rút gọn biểu thức: A = 1/a(a - b)(a - c) +1/(b(b - a)(b - c)) +1/c(c - a)(c - b)

Hướng dẫn giải

Ta có: \(A = \frac{1}{{a\left( {a - b} \right)\left( {a - c} \right)}} + \frac{1}{{b\left( {b - a} \right)\left( {b - c} \right)}} + \frac{1}{{c\left( {c - a} \right)\left( {c - b} \right)}}\)

\( = \frac{{bc\left( {b - c} \right) - ac\left( {a - c} \right) + ab\left( {a - b} \right)}}{{abc\left( {a - b} \right)\left( {b - c} \right)\left( {a - c} \right)}}\)

\( = \frac{{bc\left( {b - c} \right) - ac\left( {a - c} \right) + ab\left( {a - b} \right)}}{{abc\left( {a - b} \right)\left( {b - c} \right)\left( {a - c} \right)}}\)

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

\( = \frac{{c\left[ {\left( {b - a} \right)\left( {b + a} \right) - c\left( {b - a} \right)} \right] + ab\left( {a - b} \right)}}{{abc\left( {a - b} \right)\left( {b - c} \right)\left( {a - c} \right)}}\)

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

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

\( = \frac{{\left( {a - b} \right)\left( {a - c} \right)\left( {b - c} \right)}}{{abc\left( {a - b} \right)\left( {b - c} \right)\left( {a - c} \right)}}\)

\( = \frac{1}{{abc}}.\)

Vậy \(A = \frac{1}{{abc}}.\)