\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\)\(a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\)\(\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow\)\(\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\)\(\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\)\(\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2-3ab\right]=0\)
Do \(a+b+c\ne0\) nên \(\left(a+b\right)^2-c\left(a+b\right)+c^2-3ab=0\)
\(\Leftrightarrow\)\(a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow\)\(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\)\(\left(a^2-2ab+b^2\right)+\left(b^2-bc+c^2\right)+\left(c^2-ca+a^2\right)=0\)
\(\Leftrightarrow\)\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\)\(\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow a=b=c}\)
\(\Rightarrow\)\(N=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)
...
\(M=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)
\(M=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left(a+b\right)^2-6a^2b^2+6a^2b^2\)
\(M=\left(a+b\right)\left(a^2-ab+b^2+3a^2b+3ab^2\right)\)
\(M=\left(a^2+2ab+b^2\right)+\left(3a^2b+3ab^2-3ab\right)\)
\(M=\left(a+b\right)^2+3ab\left(a+b-1\right)\)
\(M=1+3ab.0=1\)
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Quân:Xem lại cái đề giúp t cái!
\(N=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}\)
Do \(a^3+b^3+c^3=3abc\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\hept{\begin{cases}a=b=c\\a+b+c=0\end{cases}}\). Theo đề bài thì \(a+b+c\ne0\).Do đó:
a = b = c.
Thay b và c bởi a (do a = b = c),ta có: \(N=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{3}{9}=\frac{1}{3}\)
À nhầm,không để ý là m đang trả lời câu hỏi thứ 2,sorry.
M = a3 + b3 + 3ab(a2 + b2) + 6a2b2(a + b)
= (a + b)(a2 - ab + b2) + 3ab((a + b)2 - 2ab) + 6a2b2(a + b)
= (a + b)((a + b)2 - 3ab) + 3ab((a + b)2 - 2ab) + 6a2b2(a + b)
= 1 - 3ab + 3ab(1 - 2ab) + 6a2b2
= 1 - 3ab + 3ab - 6a2b2 + 6a2b2 = 1
Một cách khác nè bn
Từ: 1a1a + 1b1b + 1c1c = 0
=> 1a31a3 + 1b31b3 + 1c31c3 = 3. 1abc1abc
Ta có:
abc2abc2 + bca2bca2 + acb2acb2
= abcc3abcc3 + abca3abca3 + abcb3abcb3
= abc. (1a31a3 + 1b31b3 + 1c31c3)
= abc. 3. 1abc1abc = 3
