Cho a,b,c là 3 số khác 0 thoả mãn (a+b+c)^2=a^2+b^2+c^2 Cmr: 1/a^3+1/b^3+1/c^3=3/abc 10/09/2021 Bởi Allison Cho a,b,c là 3 số khác 0 thoả mãn (a+b+c)^2=a^2+b^2+c^2 Cmr: 1/a^3+1/b^3+1/c^3=3/abc
Áp dụng \(\left(x+y+z\right)^3=x^3+y^3+z^3+\left(x+y+z\right)\left(xy+yz+zx\right)-3xyz\) Ta có: \(\left(a+b+c\right)^2=a^2+b^2+c^2\) => \(2ab+2ac+2bc=0\) => \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\) Khi đó: \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^3=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)-\frac{3}{abc}\) => \(0=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+0-\frac{3}{abc}\) => \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\) Bình luận
Bổ đề: \[\begin{array}{l} {x^3} + {y^3} + {z^3} – 3xyz\\ = {(x + y)^3} – 3xy(x + y) + {z^3} – 3xyz\\ = \left[ {{{(x + y)}^3} + {z^3}} \right] – 3xy(x + y) – 3xyz\\ = (x + y + z)\left[ {{{(x + y)}^2} – (x + y)z + {z^2}} \right] – 3xy(x + y + z)\\ = (x + y + z)({x^2} + {y^2} + {z^2} – xy – yz – zx) \end{array}\] \[\begin{array}{l} gt \to {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca = {a^2} + {b^2} + {c^2}\\ \to ab + bc + ca = 0\\ \frac{1}{{{a^3}}} + \frac{1}{{{b^3}}} + \frac{1}{{{c^3}}} – \frac{3}{{abc}} = (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})(\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} – \frac{1}{{ab}} – \frac{1}{{bc}} – \frac{1}{{ca}}) = (\frac{{ab + bc + ca}}{{abc}})(\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} – \frac{1}{{ab}} – \frac{1}{{bc}} – \frac{1}{{ca}}) = 0\\ \to \frac{1}{{{a^3}}} + \frac{1}{{{b^3}}} + \frac{1}{{{c^3}}} = \frac{3}{{abc}} \end{array}\] Bình luận
Áp dụng
\(\left(x+y+z\right)^3=x^3+y^3+z^3+\left(x+y+z\right)\left(xy+yz+zx\right)-3xyz\)
Ta có:
\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
=> \(2ab+2ac+2bc=0\)
=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
Khi đó:
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^3=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)-\frac{3}{abc}\)
=> \(0=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+0-\frac{3}{abc}\)
=> \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
Bổ đề:
\[\begin{array}{l}
{x^3} + {y^3} + {z^3} – 3xyz\\
= {(x + y)^3} – 3xy(x + y) + {z^3} – 3xyz\\
= \left[ {{{(x + y)}^3} + {z^3}} \right] – 3xy(x + y) – 3xyz\\
= (x + y + z)\left[ {{{(x + y)}^2} – (x + y)z + {z^2}} \right] – 3xy(x + y + z)\\
= (x + y + z)({x^2} + {y^2} + {z^2} – xy – yz – zx)
\end{array}\]
\[\begin{array}{l}
gt \to {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca = {a^2} + {b^2} + {c^2}\\
\to ab + bc + ca = 0\\
\frac{1}{{{a^3}}} + \frac{1}{{{b^3}}} + \frac{1}{{{c^3}}} – \frac{3}{{abc}} = (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})(\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} – \frac{1}{{ab}} – \frac{1}{{bc}} – \frac{1}{{ca}}) = (\frac{{ab + bc + ca}}{{abc}})(\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} – \frac{1}{{ab}} – \frac{1}{{bc}} – \frac{1}{{ca}}) = 0\\
\to \frac{1}{{{a^3}}} + \frac{1}{{{b^3}}} + \frac{1}{{{c^3}}} = \frac{3}{{abc}}
\end{array}\]