- Phân tích ra nhân tử :

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

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

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

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

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

Từ đây ta có \(A=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)}{a^2+b^2+c^2-ab-bc-ac}\)

\(\Rightarrow A=a+b+c\)

\(\frac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-bc-ca}\)

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

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

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

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

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

\(=a+b+c\)

a, Gợi ý nà :3

a^2 + b^2 - c^2 +2ab = (a^2 + b^2 + 2ab) -c^2 = (a+b)^2 - c^2 = (a + b - c)(a + b + c)

a^2 - b^2 + c^2 + 2ac = (a + c)^2 - b^2 = (a + b + c)(a - b + c)

b. Gợi ý tiếp luôn nà :3

a^3 + b^3 + c^3 - 3abc

= (a^3 + b^3 +3a^2 x b + 3ab^2) - 3ab(a+b) -3abc + c^3

= (a+b)^3 + c^3 - 3ab(a+b+c) 

= (a + b+ c)[(a+b)^2 - c(a+b) +c^2] - 3ab(a+b+c)

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

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

Rồi cứ thế rút gọn...

Học tốt nha bạn :3

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

\(\text{nhận xét: ta có hằng đẳng thức:}\)

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

đó đến đây bạn làm tiếp

b/\((\sum a^3)-3abc=(\sum a).(\sum a^2-\sum ab)\)\(\Rightarrow\)\(\frac{(\sum a^3)-3abc}{(\sum a^2-\sum ab)}=\frac {(\sum a).(\sum a^2-\sum ab)}{(\sum a^2-\sum ab)}=a+b+c\)

Tử \(a^3+b^3+c^3-3abc\)

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

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

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

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

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

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

Ta có :

\(a^3+b^3+c^3-3abc\)

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

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\) Thay vào biểu thức ta được:

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

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

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

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

=a+b+c

b: 

Sửa đề: \(=\dfrac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)

\(=\dfrac{\left(x-y\right)^3+z^3+3xy\left(x-y\right)+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{\left(x-y+z\right)\left(x^2-2xy+y^2-xz+yz+z^2\right)+3xy\left(x-y+z\right)}{2\left(x^2+y^2+z^2+xy+yz-xz\right)}\)

\(=\dfrac{\left(x-y+z\right)\left(x^2+y^2+z^2+xy-xz+yz\right)}{2\left(x^2+y^2+z^2+xy+yz-xz\right)}\)

\(=\dfrac{x-y+z}{2}\)

a) \(\dfrac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-bc-ca}\)

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

\(=a+b+c\)

Hằng đẳng thức:\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

Khi đó:

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

\(=a+b+c\)

\(=2011\)

Lời giải:

$A=a-\frac{ac}{c+a^2}+b-\frac{ab}{a+b^2}+c-\frac{bc}{b+c^2}$

$=\sum a-\sum \frac{ac}{c+a^2}$

Áp dụng BĐT AM-GM: $c+a^2\geq 2a\sqrt{c}$

$\Rightarrow A\geq \sum a-\frac{1}{2}\sum \sqrt{c}$

Áp dụng BĐT Cauchy-Schwarz:

$(\sum \sqrt{c})^2\leq (c+a+b)(1+1+1)$

$\Rightarrow \sum \sqrt{c}\leq 3\sum a$

Do đó $A\geq \sum a-\frac{1}{2}\sqrt{3\sum a}$

Đặt $\sqrt{3\sum a}=t$ thì $A\geq \frac{t^2}{3}-\frac{t}{2}(*)$

Từ điều kiện $ab+bc+ac=3abc\Rightarrow 3=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

Áp dụng BĐT Cauchy-Schwarz:

$3=\sum \frac{1}{a}\geq \frac{9}{\sum a}\Rightarrow \sum a\geq 3$

$\Rightarrow t=\sqrt{3\sum a}\geq 3$

Do đó:

$\frac{t^2}{3}-\frac{t}{2}=(t-3)(\frac{t}{3}+\frac{1}{2})+\frac{3}{2}\geq \frac{3}{2}$ với mọi $t\geq 3(**)$

Từ $(*); (**)\Rightarrow A\geq \frac{3}{2}$

Vậy $A_{\min}=\frac{3}{2}$ khi $a=b=c=1$