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

\(B=\dfrac{a^3+c^3+3ac\left(a+c\right)-b^3-3ac\left(a+c\right)+3abc}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)

\(=\dfrac{\left(a+c\right)^3-b^3-3ac\left(a+c-b\right)}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)

\(=\dfrac{\left(a+c-b\right)\left[\left(a+c\right)^2+b\left(a+c\right)+b^2\right]-3ac\left(a+c-b\right)}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)

\(=\dfrac{\left(a+c-b\right)\left(a^2+b^2+c^2+ab+bc-ac\right)}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)

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

\(=\dfrac{-2\left[\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2\right]}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}=-2\)

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)

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

Câu 1: 

a: \(\left(a+b\right)^3-3ab\left(a+b\right)\)

\(=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2\)

\(=a^3+b^3\)

b: \(a^3+b^3+c^3-3abc\)

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

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)

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

Đáp số bằng 0 nhìn ra tớ cũng biết

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To biet lam nhung ghi dai qua

Ta có: \(a^3+b^3+c^3=3abc\)

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

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

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left[a^2+2ab+b^2-ac-bc+c^2-3ab\right]=0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)

Ta có: \(N=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)

\(=\dfrac{a+b}{b}\cdot\dfrac{b+c}{c}\cdot\dfrac{a+c}{a}\)

Trường hợp 1: a+b+c=0

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)

\(\Leftrightarrow N=\dfrac{-c}{b}\cdot\dfrac{-a}{c}\cdot\dfrac{-b}{a}=\dfrac{-\left(a\cdot b\cdot c\right)}{a\cdot b\cdot c}=-1\)

Trường hợp 2: a=b=c

\(\Leftrightarrow N=\dfrac{b+b}{b}\cdot\dfrac{a+a}{a}\cdot\dfrac{c+c}{c}=2\cdot2\cdot2=8\)

1, Ta có a^3+b^3+c^3=3abc

-> a^3+b^3+c^3+3a^2b+3ab^2=3abc+3a^2b+3ab^2

-> (a+b)^{3 +} c^3 - 3ab(a+b+c)=0

-> (a+b+c). ((a+b)^2-(a+b).c+c^2)-3ab(a+b+c)=0

-> (a+b+c)(a^2+2ab+b^2-ac-bc+c^2-3ab)=0

Th1: a+b+c=0

->P= a+b/2 . b+c/2 . c+a/2

= (-c)(-a)(-b)/2=-1

TH2 a^2+b^2+c^2-ab-bc-ca=0

->2a^2+2b^2+2c^2-2ab-abc-2ac=0

->(a^2-2ab+b^2)+(a^2-2ac+c^2)+(b^2-2bc+c^2)=0

-> (a-b)^2+(a-c)^2+(b-c)^2=0

Mà (a-b)^2+(a-c)^2+(b-c)^2>= 0

Dấu = xảy ra (=)a-b=0

                         b-c=0

                          a-c=0

-> a=b=c

->P= 1+a/b+1+b/c+1+c/a=2+2+2= 8

Tại sao ko có ai làm hộ mình

\(A=\dfrac{\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc}{2\left(a^2+b^2+c^2-ab-ac-bc\right)}\)

\(=\dfrac{\left(a+b+c\right)\cdot\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)}{2\left(a^2+b^2+c^2-ab-ac-bc\right)}\)

\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)}{2\left(a^2+b^2+c^2-ab-ac-bc\right)}\)

=3/2