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ta có: (a+b+c)2 = a2 + b2 + c2
=> 2.(ab+ac+bc) = 0
ab + ac + bc = 0
=> 1/a + 1/b + 1/c = 0
Lại có: \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}-\frac{3}{abc}=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right).\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{1}{ab}-\frac{1}{ac}-\frac{1}{bc}\right).\)
\(=0.\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{1}{ab}-\frac{1}{ac}-\frac{1}{bc}\right)=0\)
=> 1/a3 + 1/b3 + 1/c3 -3/abc = 0
=> 1/a3 + 1/b3 + 1/c3 = 3/abc
Từ đkđb
\(\Leftrightarrow2\left(ab+bc+ac\right)=0\)
\(\Leftrightarrow\dfrac{ab+bc+ac}{abc}=0\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=-\dfrac{1}{c}\)
\(\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=-\dfrac{1}{c^3}\)
\(\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
Hớ hớ bài này mình cũng làm rồi.
Ta có: (a+b+c)2=a2+b2+c2
<=> a2+b2+c2+2(ab+bc+ca)=a2+b2+c2
<=>2(ab+bc+ca)=0
<=>ab+bc+ca=0
\(\Leftrightarrow\dfrac{ab+bc+ca}{abc}=0\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
=>\(\dfrac{1}{a}+\dfrac{1}{b}=-\dfrac{1}{c}\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3=\left(-\dfrac{1}{c}\right)^3\)
=> \(\dfrac{1}{a^3}+\dfrac{3}{a^2b}+\dfrac{3}{ab^2}+\dfrac{1}{b^3}=-\dfrac{1}{c^3}\)
=>\(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=-\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=-\dfrac{3}{ab}.\left(-\dfrac{1}{c}\right)=\dfrac{3}{abc}\)
=> Đpcm.
2.3+3.(-1,2)+(-1,2).2=0 (a=2, b=3, c=-1,2)
\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=\dfrac{19}{18}\)
\(\dfrac{3}{abc}=-\dfrac{5}{12}\)?
nhầm làm lại nha ^^
(a+b+c)^2=a^2+b^2+c^2
=>a^2+b^2+c^2+2ab+2bc+2ac=a^2+b^2+c^2
=>2(ab+bc+ac)=0
=>ab+bc+ac=0
=>(ab+bc+ac)/abc=0
=>ab/abc+bc/abc+ac/abc=0
=>1/c+1/a+1/b=0
=> 1/a+1/b=-1/c
=> (1/a+1/b)^3=(-1/c)^3
=> 1/a^3+1/b^3+3/ab(1/a+1/b)=-1/c^3
=> 1/a^3+1/b^3+1/c^3+3/ab.(-1/c)=0
=> 1/a^3+1/b^3+1/c^3-3/abc=0
=> 1/a^3+1/b^3+1/c^3=3/abc (đpcm)
(a+b+c)^2=a^2+b^2+c^2
a^2+b^2+c^2+2ab+2bc+2ac=a^2+b^2+c^2
2(ab+bc+ac)=0
ab+bc+ac=0
(ab+bc+ac)/abc=0
ab/abc+bc/abc+ac/abc=0
1/c+1/a+1/b=0
=> 1/a+1/b=-1/c
=> (1/a+1/b)^3=(-1/c)^3
=> 1/a^3+1/b^3+3.(1/a.)(1/b).(1/a+1/b)=-1/c^3
=> 1/a^3+1/b^3+1/c^3.3ab.(-1/c)=0
=> 1/a^3+1/b^3+1/c^3=3/abc