Cho 1/a + 1/b + 1/c = 0 (abc khác 0). Tính biểu thức A = (b + c) / a + (c + a) / b
Cho: \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0\) (abc ≠ 0). Tính biểu thức: \(A = \frac{{b + c}}{a} + \frac{{c + a}}{b} + \frac{{a + b}}{c}\).
Cho: \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0\) (abc ≠ 0). Tính biểu thức: \(A = \frac{{b + c}}{a} + \frac{{c + a}}{b} + \frac{{a + b}}{c}\).
Ta có: \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0\)
\( \Leftrightarrow \frac{{ab + bc + ca}}{{abc}} = 0\)
⇔ ab + bc + ca = 0
Mặt khác, \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0\)
\( \Leftrightarrow \frac{1}{a} + \frac{1}{b} = - \frac{1}{c}\)
\( \Leftrightarrow {\left( {\frac{1}{a} + \frac{1}{b}} \right)^3} = - \frac{1}{{{c^3}}}\)
\( \Leftrightarrow \frac{1}{{{a^3}}} + \frac{1}{{{b^3}}} + 3.\frac{1}{{ab}}.\left( {\frac{1}{a} + \frac{1}{b}} \right) = - \frac{1}{{{c^3}}}\)
\( \Leftrightarrow \frac{1}{{{a^3}}} + \frac{1}{{{b^3}}} + 3.\frac{1}{{ab}}.\left( { - \frac{1}{c}} \right) = \frac{{ - 1}}{{{c^3}}}\)
\( \Leftrightarrow \frac{1}{{{a^3}}} + \frac{1}{{{b^3}}} + \frac{1}{{{c^3}}} = \frac{3}{{abc}}\) (*)
Khi đó: \(\frac{{\left( {b + c} \right)}}{a} = \frac{{ab + ac}}{{{a^2}}} = \frac{{ - bc}}{{{a^2}}} = \frac{{ - abc}}{{{a^2}}}\)
Tương tự ta có: \(\frac{{\left( {a + b} \right)}}{c} = \frac{{ - abc}}{{{c^3}}}\); \(\frac{{\left( {a + c} \right)}}{{{b^2}}} = \frac{{ - abc}}{{{b^3}}}\).
\(M = \frac{{ - abc}}{{{a^3}}} + \frac{{ - abc}}{{{b^3}}} + \frac{{ - abc}}{{{c^3}}}\)
\( = - abc\left( {\frac{1}{{{a^3}}} + \frac{1}{{{b^3}}} + \frac{1}{{{c^3}}}} \right)\)
\( = - abc.\frac{3}{{abc}} = - 3\) (theo *)
Vậy M = −3.