Đáp số bằng 0 nhìn ra tớ cũng biết

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Đầu tiên bạn hãy tự phân tích tử số nha, kết quả là:

    \(a^3+b^3+c^3-3abc=\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\)

Ta có: \(a+b+c=3\)

Vậy thay vào biểu thức, ta sẽ được:

    \(S=\frac{a^3+b^3+c^3-3abc}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)

\(\Leftrightarrow S=\frac{\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)

\(\Leftrightarrow S=\frac{1}{2}\left(a+b+c\right)\)

\(\Leftrightarrow S=\frac{1}{2}.3\)

\(\Leftrightarrow S=\frac{3}{2}\)

Chúc bạn học giỏi và tíck cho mìk vs nha Đỗ Nguyễn Hiền Thảo!

\(a^3+b^3+c^3=3abc\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)

\(\Leftrightarrow\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)

\(\Rightarrow\orbr{\begin{cases}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\end{cases}\Rightarrow\orbr{\begin{cases}a+b+c=0\\a=b=c\end{cases}}}\)

Với \(a+b+c=0\) thì \(\hept{\begin{cases}a+b=-c\\a+c=-b\\b+c=-a\end{cases}}\)

\(\Rightarrow A=\frac{a+b}{b}.\frac{b+c}{c}.\frac{a+c}{a}=\frac{-c}{b}.\frac{-a}{c}.\frac{-b}{a}=-1\)

Với \(a=b=c\) thì :

\(A=\left(1+\frac{a}{a}\right)\left(1+\frac{b}{b}\right)\left(1+\frac{c}{c}\right)=2.2.2=8\)

phân tích tử thức: 

\(a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

Phân tích mẫu thức:\(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(ab^2-a^2b+bc^2-b^2c+ca^2-c^2a\right)\)

\(=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

\(\Rightarrow A=\frac{3\left(a^2+b^2+c^2-ab-bc-ca\right)}{3\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{a^2+b^2+c^2-ab-bc-ca}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

a 3 + b 3 + c 3 = 3abc⇔a 3 + b 3 + c 3 − 3abc = 0

⇔ a + b 3 − 3ab a + b + c 3 − 3abc = 0

⇔ a + b 3 + c 3 − 3ab a + b + 3abc = 0

⇔ a + b + c a 2 + b 2 + c 2 + 2ab − ac − bc − 3ab a + b + c = 0

⇔ a + b + c a 2 + b 2 + c 2 − ab − bc − ac = 0

⇔ 2 a + b + c a − b 2 + b − c 2 + c − a /2 = 0

Vì a,b,c > 0 nên a+b+c > 0

Do đó : a − b 2 = 0

             b − c 2 = 0 

             c − a 2 = 0

⇒a = b = c

k cho mk nha

\(A=\frac{a^3+b^3+c^3-3abc}{\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3}.\)

Áp dụng: (a+b+c)³=a³+b³+c³+3(a+b)(b+c)(c+a) => a³+b³+c³=(a+b+c)³-3(a+b)(b+c)(c+a)=27-3(a+b)(b+c)(c+a)

=> \(A=\frac{27-3\left(a+b\right)\left(b+c\right)\left(c+a\right)-3abc}{a^3-3a^2b+3ab^2-b^3+b^3-3b^2c+3bc^2-c^3+c^3-3c^2a+3ca^2-a^3}.\)

\(A=\frac{27-3\left(a+b\right)\left(b+c\right)\left(c+a\right)-3abc}{-3a^2b+3ab^2-3b^2c+3bc^2-3c^2a+3ca^2}\)=> \(A=\frac{9-\left(a+b\right)\left(b+c\right)\left(c+a\right)-abc}{-a^2b+ab^2-b^2c+bc^2-c^2a+ca^2}\)

Ta có: (a+b)(b+c)(c+a)=(3-c)(3-b)(3-a)=27-9a-9b-9c+3ab+3ac+3bc-abc=27-9(a+b+c)+3(ab+bc+ca)-abc=3(ab+bc+ca)-abc

Và: -a²b+ab²-b²c+bc²-c²a+ca²=(a-b)(b-c)(c-a)

=> \(A=\frac{9-3\left(ab+bc+ca\right)+abc-abc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(A=\frac{9-3\left(ab+bc+ca\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

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