Toán Cho ab+bc+ca=0 và a,b,c khác 0 CM (3a^3-bc)/a^3bc=1/b^3+1/c^3 31/07/2021 By Caroline Cho ab+bc+ca=0 và a,b,c khác 0 CM (3a^3-bc)/a^3bc=1/b^3+1/c^3
Giải thích các bước giải: Ta có: \[\begin{array}{l}ab + bc + ca = 0 \Leftrightarrow \frac{{ab + bc + ca}}{{abc}} = 0 \Leftrightarrow \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0\\\frac{1}{{{a^3}}} + \frac{1}{{{b^3}}} + \frac{1}{{{c^3}}} – \frac{3}{{abc}} = {\left( {\frac{1}{a} + \frac{1}{b}} \right)^3} – 3.\frac{1}{a}.\frac{1}{b}.\left( {\frac{1}{a} + \frac{1}{b}} \right) + \frac{1}{{{c^3}}} – \frac{3}{{abc}}\\ = \left[ {{{\left( {\frac{1}{a} + \frac{1}{b}} \right)}^3} + \frac{1}{{{c^3}}}} \right] – \frac{3}{{ab}}\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right)\\ = {\left( {\frac{1}{a} + \frac{1}{b}} \right)^3} + \frac{1}{{{c^3}}}\\ = \left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right)\left( {{{\left( {\frac{1}{a} + \frac{1}{b}} \right)}^2} – \left( {\frac{1}{a} + \frac{1}{b}} \right).\frac{1}{c} + \frac{1}{{{c^2}}}} \right)\\ = 0\\ \Rightarrow \frac{1}{{{b^3}}} + \frac{1}{{{c^3}}} = \frac{3}{{abc}} – \frac{1}{{{a^3}}}\\ \Leftrightarrow \frac{1}{{{b^3}}} + \frac{1}{{{c^3}}} = \frac{{3{a^2} – bc}}{{{a^3}bc}}\end{array}\] Trả lời
Giải thích các bước giải:
Ta có:
\[\begin{array}{l}
ab + bc + ca = 0 \Leftrightarrow \frac{{ab + bc + ca}}{{abc}} = 0 \Leftrightarrow \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0\\
\frac{1}{{{a^3}}} + \frac{1}{{{b^3}}} + \frac{1}{{{c^3}}} – \frac{3}{{abc}} = {\left( {\frac{1}{a} + \frac{1}{b}} \right)^3} – 3.\frac{1}{a}.\frac{1}{b}.\left( {\frac{1}{a} + \frac{1}{b}} \right) + \frac{1}{{{c^3}}} – \frac{3}{{abc}}\\
= \left[ {{{\left( {\frac{1}{a} + \frac{1}{b}} \right)}^3} + \frac{1}{{{c^3}}}} \right] – \frac{3}{{ab}}\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right)\\
= {\left( {\frac{1}{a} + \frac{1}{b}} \right)^3} + \frac{1}{{{c^3}}}\\
= \left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right)\left( {{{\left( {\frac{1}{a} + \frac{1}{b}} \right)}^2} – \left( {\frac{1}{a} + \frac{1}{b}} \right).\frac{1}{c} + \frac{1}{{{c^2}}}} \right)\\
= 0\\
\Rightarrow \frac{1}{{{b^3}}} + \frac{1}{{{c^3}}} = \frac{3}{{abc}} – \frac{1}{{{a^3}}}\\
\Leftrightarrow \frac{1}{{{b^3}}} + \frac{1}{{{c^3}}} = \frac{{3{a^2} – bc}}{{{a^3}bc}}
\end{array}\]