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http://diendantoanhoc.net/topic/152549-t%C3%ADnh-fraca2a2-b2-c2-fracb2b2-c2-a2fracc2c2-b2-a2/
Ta có: \(a+b+c=0\)
\(\Rightarrow1\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a^2+b^2=-2ab+c^2\\b^2+c^2=-2bc+a^2\\c^2+a^2=-2ac+b^2\end{cases}}\)
\(\Rightarrow1A=\frac{a^2}{a^2+2bc-a^2}+\frac{b^2}{b^2+2ac-b^2}+\frac{c^2}{c^2+2ab-c^2}\)
\(=\frac{a^3+b^3+c^3}{2abc}=\frac{a^3+b^3+c^3-3abc+3abc}{2abc}\)
\(=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc}{2abc}\)
\(=\frac{3}{2}\)
Ta có: \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0\)
\(\Rightarrow\frac{bcx+acy+abz}{abc}=0\)
\(\Rightarrow bcx+acy+abz=0\)
Lại có:\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2\)
\(\Rightarrow\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}+2.\frac{bcx+acy+abz}{xyz}=4\)(bình phương hai vế)
\(\Rightarrow\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}=4\)(Vì \(bcx+acy+abz=0\))
Từ (1) \(\Rightarrow bcx+acy+abz=0\)
Gọi \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2\left(2\right)\)
Từ (2) \(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{ab}{xy}+\frac{ac}{xz}+\frac{bc}{yz}\right)=0\)
\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=4-\left(\frac{abz+acy+bcx}{xyz}\right)\)
\(=4\)
\(b,\frac{ab}{a^2+b^2+c^2}+\frac{bc}{b^2+c^2-a^2}+\frac{ca}{c^2+a^2-b^2}\)
Từ \(a+b+c=0\Rightarrow a+b=-c\Rightarrow a^2+b^2-c^2=-2ab\)
Tương tự \(b^2+c^2-a^2=-2bc\)và \(c^2+a^2-b^2=-2ac\)
\(\Rightarrow\frac{ab}{-2ab}+\frac{bc}{-2bc}+\frac{ca}{-2ca}=\frac{1}{-2}+\frac{1}{-2}+\frac{1}{-2}\)
\(=-\frac{3}{2}\)
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:
\(P=\frac{-\left(3abc\right)}{2abc}=-\frac{3}{2}\)
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:
∑\(\frac{a^2}{a^2-b^2-c^2}=\)∑\(\frac{a^2}{2bc}=\frac{a^3+b^3+c^3}{2abc}=\frac{3abc}{2abc}=\frac{3}{2}\)