Lời giải:
\(-A=\frac{a^2}{(a-b)(c-a)}+\frac{b^2}{(a-b)(b-c)}+\frac{c^2}{(c-a)(b-c)}\)

\(=\frac{a^2(b-c)+b^2(c-a)+c^2(a-b)}{(a-b)(b-c)(c-a)}=\frac{a^2b+b^2c+c^2a-(ab^2+bc^2+ca^2)}{-[(a^2b+b^2c+c^2a)-(ab^2+bc^2+ca^2)]}=-1\)

$\Rightarrow A=1$

a) A = \(\frac{a}{\left(a-b\right)\left(a-c\right)}+\frac{b}{\left(b-a\right)\left(b-c\right)}+\frac{c}{\left(c-a\right)\left(c-b\right)}\)

=> A = \(\frac{a}{\left(a-b\right)\left(a-c\right)}-\frac{b}{\left(a-b\right)\left(b-c\right)}+\frac{c}{\left(a-c\right)\left(b-c\right)}\)

=> A = \(\frac{a\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}-\frac{b\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{c\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

=> A + \(\frac{ab-ac-ab+bc+ac-bc}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=0\)

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

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

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

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

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

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

\(=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)

Đặt \(\hept{\begin{cases}a-b=x\\b-c=y\\c-a=z\end{cases}}\)

\(A=\frac{2}{x}+\frac{2}{y}+\frac{2}{z}+\frac{x^2y^2z^2}{xyz}\)

\(A=\frac{\left(2y+2x\right).z+2xy}{xyz}+\frac{x^2+y^2+x^2}{xyz}\)

\(A=\frac{2yz+2xz+2xy}{xyz}+\frac{x^2+y^2+z^2}{xyz}\)

\(A=\frac{2yz+2xz+2xy+x^2+y^2+z^2}{xyz}=\frac{\left(x+y+z\right)^2}{xyz}\)

Có đúng k nhỉ k chắc

ta có: \(T=\frac{a^2}{\left(a-b\right).\left(a+b\right)-c^2}+\frac{b^2}{\left(b-c\right).\left(b+c\right)-a^2}+\frac{c^2}{\left(c-a\right).\left(c+a\right)-b^2}\)

\(T=\frac{a^2}{a^2-b^2-c^2}+\frac{b^2}{b^2-c^2-a^2}+\frac{c^2}{c^2-a^2-b^2}\)

mà a + b + c = 0 => b + c = -a => b² + 2bc + c² = a²  => a² - b² - c² = 2bc

tương tự như trên, ta có: b²  - c² - a² = 2ac; c² - a² - b² = 2ab

\(\Rightarrow T=\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}=\frac{a^3+b^3+c^3}{2abc}\)

Lại có: a+b+c = 0 => a³ + b³ + c³ = 3abc

\(\Rightarrow T=\frac{3abc}{2abc}=\frac{3}{2}\)

Bài cuối hơi khó nhìn, bạn thông cảm nhé! ^^

a) \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)

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

\(=\left(a^2b-c^2b\right)-\left(a^2c-c^2a\right)-b^2\left(a-c\right)\)

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

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

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

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

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

\(=\left(a-c\right)\left[b\left(a-b\right)+c\left(b-a\right)\right]\)

\(=\left(a-c\right)\left[b\left(a-b\right)-c\left(a-b\right)\right]\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(=\left(a-c\right)\left(b-c\right)\left[\left(a-b\right)\left(a+b\right)+c\left(a-b\right)\right]\)

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

Khúc cuối nè

\(Ta\) \(có:\) \(1+a^2=ab+bc+ca+a^2=b\left(a+c\right)+a\left(a+c\right)=\left(a+b\right)\left(c+a\right)\)

\(1+b^2=ab+bc+ca+b^2=\left(a+b\right)\left(b+c\right)\)

\(1+c^2=ab+bc+ca+c^2=\left(a+c\right)\left(c+b\right)\)

\(Khi\) \(đó:\) \(A=\dfrac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(c+b\right)}\)

\(\Rightarrow A=1\)

a) Ta có : \(a^2+1=a^2+ab+bc+ac=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)

Tương tự : \(b^2+1=\left(b+a\right)\left(b+c\right)\) ; \(c^2+1=\left(c+a\right)\left(c+b\right)\)

Suy ra \(\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)=\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\)

Vậy \(A=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}=1\)

b) Ta có ; \(a^2+2bc-1=a^2+2bc-\left(ab+bc+ac\right)=a^2-ab+bc-ac=a\left(a-b\right)-c\left(a-b\right)\)

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

Tương tự : \(b^2+2ac-1=\left(a-b\right)\left(c-b\right)\) ; \(c^2+2ab-1=\left(a-c\right)\left(b-c\right)\)

Suy ra \(\left(a^2+2bc-1\right)\left(b^2+2ac-1\right)\left(c^2+2ab-1\right)=\left(a-b\right)^2.\left(c-a\right)^2.\left[-\left(b-c\right)^2\right]\)

Vậy : \(B=\frac{-\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)}=-1\)