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Ta có:
\(a^3+b^3+c^3=3abc=>a^3+b^3+c^3-3abc=0\)
\(=>\left(a+b\right)^3-3a^2b-3ab^2+c^3-3abc=0\)
\(=>\left[\left(a+b\right)^3+c^3\right]-3a^2b-3ab^2-3abc=0\)
\(=>\left[\left(a+b\right)^3+c^3\right]-3ab\left(a+b+c\right)=0\)
\(=>\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
\(=>\left(a+b+c\right)\left(a^2+2ab+b^2-ca-bc+c^2-3ab\right)=0\)
\(=>\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
Vì a3+b3+c3=3abc và a+b+c khác 0
=>\(a^2+b^2+c^2-ab-bc-ca=0\)
\(=>2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(=>\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(=>\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Tổng 3 số không âm = 0 <=> chúng đều = 0
\(< =>\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}< =>a=b=c}\)
Vậy \(\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{1}{3}\)
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Ta có ; \(a^3+b^3+c^3=3abc\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-bc-ac\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\frac{a+b+c}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
Vì \(a+b+c\ne0\) nên ta có \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\Leftrightarrow a=b=c\)
a) Thay a = b = c vào biểu thức được : \(\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)
b) Thay a = b = c vào P : \(P=\frac{2}{a}.\frac{2}{b}\frac{2}{c}=\frac{8}{abc}\)
\(P=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-a\right)\left(b-c\right)}+\frac{c^2}{\left(c-b\right)\left(c-a\right)}\)
\(P=\frac{a^2}{\left(a-b\right)\left(a-c\right)}-\frac{b^2}{\left(a-b\right)\left(b-c\right)}+\frac{c^2}{\left(c-b\right)\left(c-a\right)}\)
\(P=\frac{1}{a-b}.\frac{a^2\left(b-c\right)-b^2\left(a-c\right)}{\left(a-c\right)\left(b-c\right)}+\frac{c^2}{\left(c-b\right)\left(c-a\right)}\)
\(P=\frac{1}{a-b}.\frac{a^2b-a^2c-b^2a+b^2c}{\left(a-c\right)\left(b-c\right)}+\frac{c^2}{\left(c-b\right)\left(c-a\right)}\)
\(P=\frac{1}{a-b}.\frac{ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)}{\left(a-c\right)\left(b-c\right)}+\frac{c^2}{\left(c-b\right)\left(c-a\right)}\)
\(P=\frac{1}{a-b}.\frac{\left(a-b\right)\left(ab-ac-bc\right)}{\left(a-c\right)\left(b-c\right)}+\frac{c^2}{\left(b-c\right)\left(a-c\right)}\)
\(P=\frac{ab-ac-bc}{\left(a-c\right)\left(b-c\right)}+\frac{c^2}{\left(a-c\right)\left(b-c\right)}\)
\(P=\frac{ab-ac-bc+c^2}{\left(a-c\right)\left(b-c\right)}=\frac{a\left(b-c\right)-c\left(b-c\right)}{\left(a-c\right)\left(b-c\right)}=\frac{\left(a-c\right)\left(b-c\right)}{\left(a-c\right)\left(b-c\right)}\)
=> P = 1
Đáp số: P=1
\(P=-\frac{a^2}{\left(a-b\right)\left(c-a\right)}-\frac{b^2}{\left(a-b\right)\left(b-c\right)}-\frac{c^2}{\left(c-a\right)\left(b-c\right)}\)
\(=-\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=-\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)