cho a,b,c khac 0 va a+b+c=0 . tinh Q=\(\frac{1}{a^2+b^2-c^2}+\frac{1}{b^2+c^2-a^2}+\frac{1}{a^2+c^2-b^2}\)
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TN
14 tháng 11 2018
\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=a^2+b^2+c^2\)
\(\Rightarrow2\left(ab+bc+ac\right)=0\)
\(\Rightarrow ab+bc+ac=0\)
\(\Rightarrow\frac{\left(a+b+c\right)}{abc}=0\)
\(\Rightarrow\frac{ab}{abc}+\frac{bc}{abc}+\frac{ac}{abc}=0\)
\(\Rightarrow\frac{1}{c}+\frac{1}{a}+\frac{1}{b}=0\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{-1}{c}\)
\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(\frac{-1}{c}\right)^3\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=-\frac{1}{c^3}\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{3}{ab}.\left(-\frac{1}{c}\right)=0\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}-\frac{3}{ab}=0\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\left(đpcm\right)\)
14 tháng 11 2018
\(\left(a+b+c\right)^2=a^2+b^2+c^2\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=a^2+b^2+c^2\Rightarrow ab+bc+ac=0\)
\(\Rightarrow\frac{ab+bc+ac}{abc}=0\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\left(\frac{1}{a}\right)^3+\left(\frac{1}{b}\right)^3+\left(\frac{1}{c}\right)^3=3.\frac{1}{a}.\frac{1}{b}.\frac{1}{c}\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
1 tháng 1 2016
Có: (a+b+c)2=a2+b2+c2
=> a2 +b2 +c2 +2(a*b+b*c+c*a)=a2 +b2 +c2
=>2*(a*b+b*c+c*a) = 0
=>a*b+b*c+c*a = 0
=> (a*b+b*c+c*a)/a*b*c = 0 ( cùng chia 2 vế cho a*b*c)
=> (a*b/a*b*c)+(b*c/a*b*c)+(c*a/a*b*c) = 0
=>1/c+1/a+1/b = 0
=>1/a3 +1/b3 +1/c3 =3*1/a*1/b*1/c = 3/a*b*c
S
14 tháng 11 2018
Ta có: a3+b3+c3=3abc <=> a3+b3+c3-3abc=0
<=>\(a^3+3a^2b+3ab^2+b^3+c^3-3ab\left(a+b\right)-3abc=0\)
<=>\(\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
<=>\(\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
<=>\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
<=>\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
Mà a+b+c khác 0
=>\(a^2+b^2+c^2-ab-bc-ca=0\)
<=>\(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
<=>\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
<=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
<=>\(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow}}a=b=c}\)
=>\(N=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)
6 tháng 3 2020
- Ta có : \(a^3+b^3+c^3=3abc\)
=> \(a^3+b^3+c^3-3abc=0\)
=> \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
Mà \(a+b+c\ne0\)
=> \(a^2+b^2+c^2-ab-bc-ac=0\)
=> \(\frac{\left(a^2-2ab+b^2\right)+\left(b^2-2ac+c^2\right)+\left(c^2-2ac+a^2\right)}{2}=0\)
=> \(\frac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{2}=0\)
=> \(a-b=b-c=c-a=0\)
=> \(a=b=c\)
- Thay a = b = c vào biểu thức N ta được :
\(N=\frac{a^2+a^2+a^2}{\left(a+a+a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)
Vậy giá trị của N = \(\frac{1}{3}\) khi \(a^3+b^3+c^3=3abc\) và \(a+b+c\ne0\)
a + b + c = 0 => c = -a - b ; b= -a - c ; a = - b - c
Thay vào Q ta có :
\(Q=\frac{1}{a^2+b^2-\left(a+b\right)^2}+\frac{1}{b^2+c^2-\left(b+c\right)^2}+\frac{1}{a^2+c^2-\left(a+c\right)^2}\)
\(Q=\frac{1}{a^2+b^2-a^2-b^2-2ab}+\frac{1}{b^2+c^2-b^2-c^2-2bc}+\frac{1}{c^2+a^2-c^2-a^2-2ac}\)
\(Q=\frac{1}{-2ab}+\frac{1}{-2bc}+\frac{1}{-2ac}=\frac{c+a+b}{-2abc}=0\)