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\(a+b+c=0\Leftrightarrow a+b=-c\Leftrightarrow\left(a+b\right)^2=\left(-c\right)^2\Leftrightarrow a^2+b^2+2ab=c^2\Leftrightarrow a^2+b^2-c^2=-2ab\)
tương tự ta có: b2+c2-a2=-2bc ; a2+c2-b2=-2ac
Do đó \(P=\frac{1}{-2bc}+\frac{1}{-2ca}+\frac{1}{-2ab}=\frac{a+b+c}{-2abc}=0\)
P= \(\dfrac{1}{b^2+c^2-a^2}+\dfrac{1}{a^2+c^2-b^2}+\dfrac{1}{a^2+b^2-c^2}\)
=
\(\dfrac{a+b+c}{\left(b^2+c^2-a^2\right)\left(a+b+c\right)}+\dfrac{a+b+c}{\left(a^2+c^2-b^2\right)\left(a+b+c\right)}+\dfrac{a+b+c}{\left(a^2+b^2-c^2\right)\left(a+b+c\right)}\)
= 0+0+0 = 0
Vậy P= 0
Ngu vãi ko bt đúng không nx
Có \(\left\{{}\begin{matrix}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a^2=\left(b+c\right)^2\\b^2=\left(a+c\right)^2\\c^2=\left(a+b\right)^2\end{matrix}\right.\)Thay vào M đc
\(M=\frac{a^2}{2bc}+\frac{b^2}{2ca}+\frac{c^2}{2ab}\)\(\Leftrightarrow M=\frac{1}{2}\left(\frac{a^3+b^3+c^3}{abc}\right)\)
Tháy hơi sai đề rồi
Ta có : a+b+c=0\(\Rightarrow a^2=\left(b+c\right)^2\Rightarrow a^2-b^2-c^2=2bc\)
Tương tự, ta có:
∑\(\frac{a^2}{a^2-b^2-c^2}=\)∑\(\frac{a^2}{2bc}=\frac{a^3+b^3+c^3}{2abc}=\frac{3abc}{2abc}=\frac{3}{2}\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Leftrightarrow\frac{bc+ca+ab}{abc}=0\)
\(\Leftrightarrow bc+ca+ab=0\)
\(\Leftrightarrow\hept{\begin{cases}bc=-ab-ca\\ca=-ab-bc\\ab=-ca-bc\end{cases}}\)
Ta có : \(A=\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}\)
\(\Leftrightarrow A=\frac{a^2}{a^2+bc-ab-ca}+\frac{b^2}{b^2+ac-ab-bc}+\frac{c^2}{c^2+ab-ca-bc}\)
\(\Leftrightarrow A=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-a\right)\left(b-c\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)
\(\Leftrightarrow A=\frac{a^2}{\left(a-b\right)\left(a-c\right)}-\frac{b^2}{\left(b-c\right)\left(a-b\right)}+\frac{c^2}{\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{a^2\left(b-c\right)-b^2\left[\left(b-c\right)+\left(a-b\right)\right]+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{a^2\left(b-c\right)-b^2\left(b-c\right)-b^2\left(a-b\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{\left(a^2-b^2\right)\left(b-c\right)-\left(b^2-c^2\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{\left(a+b\right)\left(a-b\right)\left(b-c\right)-\left(b+c\right)\left(b-c\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{\left(a-b\right)\left(b-c\right)\left[\left(a+b\right)-\left(b+c\right)\right]}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow A=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)
http://diendantoanhoc.net/topic/152549-t%C3%ADnh-fraca2a2-b2-c2-fracb2b2-c2-a2fracc2c2-b2-a2/
Ta có: \(a+b+c=0\)
\(\Rightarrow1\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a^2+b^2=-2ab+c^2\\b^2+c^2=-2bc+a^2\\c^2+a^2=-2ac+b^2\end{cases}}\)
\(\Rightarrow1A=\frac{a^2}{a^2+2bc-a^2}+\frac{b^2}{b^2+2ac-b^2}+\frac{c^2}{c^2+2ab-c^2}\)
\(=\frac{a^3+b^3+c^3}{2abc}=\frac{a^3+b^3+c^3-3abc+3abc}{2abc}\)
\(=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc}{2abc}\)
\(=\frac{3}{2}\)
abc khác 0 nhé ạ
Do \(a+b+c=0\)
\(\Rightarrow c=-a-b\)
\(\Rightarrow c^2=a^2+2ab+b^2\)
Tương tự,ta có:
\(a^2=b^2+2bc+c^2\)
\(b^2=a^2+2ac+c^2\)
Thay vào bài toán,ta được:
\(P=\frac{c^2}{a^2+b^2-\left(a^2+2ab+b^2\right)}+\frac{a^2}{b^2+c^2-\left(b^2+2bc+c^2\right)}+\frac{b^2}{c^2+a^2-\left(a^2+2ac+c^2\right)}\)
\(P=\frac{-c^2}{2ab}+\frac{-a^2}{2bc}+\frac{-b^2}{2ac}\)
\(P=\frac{-\left(a^3+b^3+c^3\right)}{2abc}\)
Do \(a+b+c=0\Rightarrow-a=b+c\)
\(\Rightarrow-a^3=b^3+c^3+3bc\left(b+c\right)\)
\(\Rightarrow-a^3=b^3+c^3-3abc\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
Khi đó,ta có:
\(P=\frac{-\left(3abc\right)}{2abc}=-\frac{3}{2}\)