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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 \(\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{1}{3}\)
\(\)
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 : \(P=\frac{2}{a}.\frac{2}{b}\frac{2}{c}=\frac{8}{abc}\)
\(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)^3-3\left(a+b\right)c\left(a+b+c\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)[\left(a+b+c\right)^2-3ab-3ac-3bc]\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(=\frac{1}{2}\left(a+b+c\right).2\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(=\frac{1}{2}\left(a+b+c\right)[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)]\)
\(=\frac{1}{2}\left(a+b+c\right)[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2]\)
Ta có : \(a^3+b^3+c^3=3abc\)
\(\Rightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\left(1\right)\\a^2+b^2+c^2-ab-bc-ca=0\left(2\right)\end{cases}}\)
Từ (1) \(\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)
Khi đo s: \(P=\frac{abc}{\left(-a\right)\left(-b\right)\left(-c\right)}=-1\)
Từ (2) \(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow a=b=c\)
Khi đó : \(P=\frac{a^3}{2a\cdot2a\cdot2a}=\frac{1}{8}\)
Vậy : \(P=\frac{1}{8}\) hoặc \(P=-1\) với a,b,c thỏa mãn đề.
Do \(a,b,c\) là các số dương suy ra:
\(a>0;b>0;c>0\)
Suy ra: \(a+b+c>0\)
Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2-3ab\left(a+b+c\right)\right]=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Rightarrow a+b+c=0\) hoặc \(a^2+b^2+c^2-ab-bc-ca=0\)
Do \(a+b+c>0\)
Suy ra: \(a^2+b^2+c^2-ab-bc-ca=0\)
\(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Suy ra: \(a-b=0;b-c=0\) và \(c-a=0\)
Suy ra: \(a=b=c\)
Suy ra: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=1\)
Ta có: \(\left(\frac{a}{b}-1\right)+\left(\frac{b}{c}-1\right)+\left(\frac{c}{a}-1\right)=\left(1-1\right)+\left(1-1\right)+\left(1-1\right)=0\)
Vậy ...
Sau khi giải bài này xong mình cảm thấy hoa mắt và chóng mặt, mong GP sẽ gấp đôi :)
Ta có:\(A^3+B^3+C^3-3ABC=A^3+3A^2B+3AB^2+B^3+C^3-3AB\left(A+B+C\right)\)
\(=\left(A+B\right)^3+C^3-3AB\left(A+B+C\right)\)\(=\left(A+B+C\right)\left(A^2+B^2+C^2-AB-BC-CA\right)\)
Mặt khác:\(\left(A-B\right)^2+\left(B-C\right)^2+\left(C-A\right)^2=A^2-2AB+B^2+B^2-2BC+C^2+C^2-2CA+A^2\)
\(=2\left(A^2+B^2+C^2-AB-BC-CA\right)\)
Nên giá trị của phân thức là:\(\frac{A+B+C}{2}\)