Umk !!! giúp liền nàk
\(a+b+c=0\Rightarrow\hept{\begin{cases}a+b=-c\\a+c=-b\\b+c=-a\end{cases}}\)nên
\(\frac{a^2}{a^2-b^2-c^2}+\frac{b^2}{b^2-c^2-a^2}+\frac{c^2}{c^2-a^2-b^2}\)
\(=\frac{a^2}{\left(a-b\right)\left(a+b\right)-c^2}+\frac{b^2}{\left(b-c\right)\left(b+c\right)-a^2}+\frac{c^2}{\left(c-a\right)\left(c+a\right)-b^2}\)
\(=\frac{a^2}{-c\left(a-b\right)-c^2}+\frac{b^2}{-a\left(b-c\right)-a^2}+\frac{c^2}{-b\left(c-a\right)-b^2}\)
\(=\frac{a^2}{-ac+bc-c^2}+\frac{b^2}{-ab+ac-a^2}+\frac{c^2}{-bc+ab-b^2}\)
\(=\frac{a^2}{-c\left(a+c\right)+bc}+\frac{b^2}{-a\left(a+b\right)+ac}+\frac{c^2}{-b\left(b+c\right)+ab}\)
\(=\frac{a^2}{-c\left(-b\right)+bc}+\frac{b^2}{\left(-a\right)\left(-c\right)+ac}+\frac{c^2}{-b\left(-a\right)+ab}\)
\(=\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}=\frac{a^3+b^3+c^3}{2abc}\)
Mà a + b +c = 0 nên \(a^3+b^3+c^3=3abc\) (tự chứng minh)
Do đó \(\frac{a^3+b^3+c^3}{2abc}=\frac{3abc}{2abc}=\frac{3}{2}\)
Vậy \(\frac{a^2}{a^2-b^2-c^2}+\frac{b^2}{b^2-c^2-a^2}+\frac{c^2}{c^2-a^2-b^2}=\frac{3}{2}\)
trả ơn này
Vì a + b + c = 0
\(\Rightarrow\)a2 = b2 + c2 + 2bc \(\Rightarrow\) a2 - b2 - c2 = 2bc
\(\Rightarrow\)b2 = a2 + c2 + 2bc\(\Rightarrow\) b2 - a2 - c2 = 2bc
\(\Rightarrow\) c2 = a2 + c2 +2ab\(\Rightarrow\)c2 - b2 - a2 = 2ab
còn lại tự làm nhé
