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

(a+b-c)+(a+b)-(a-b-c)=a+b-c+a+b-a+b+c=a+3b

=(b+c)(ac-a²+bc-ab)+(b+c)(ac-bc+a²-ab)+(c+a)(a+b)(b-c)

=(b+c)(ac-a²+bc-ab+ac-bc+a²-ab)+(a+c)(a+b)(b-c)

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

=(b+c).2a.(c-b)-(a²+ab+ac+bc)(c-b)

=(c-b)(2ab+2ac-a²-ab-ac-bc)

=(c-b)(-a²+ab+ac-bc)=(c-b)[a(b-a)-c(b-a)]

=(c-b)(b-a)(a-c)

a: =b-c-a+c+1-a-b+c

=-2a+1

b: =a-b-c-b+c+a+c-b-a

=c-3b+a

c: =2(a-b-b+c-c+a)

=2(2a-2b)

=4a-4b

a) \(\left(b-c\right)-\left(a-c-1\right)-\left(a+b-c\right)\)

\(=b-c-a+c+1-a-b+c\)

\(=c-2a+1\)

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

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

\(=a-3b+c\)

c) \(2\cdot\left(a-b\right)-2\cdot\left(b-c\right)-2\cdot\left(c-a\right)\)

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

\(=2\cdot\left(2a-2b\right)\)

\(=4a-4b\)

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

1.

Cho $a+b+c=0$.

Ta có:

$a+b+c=0\Rightarrow a=-(b+c)$

$\Rightarrow a^2=(b+c)^2=b^2+c^2+2bc$

$\Rightarrow a^2-b^2-c^2=2bc$

Tương tự:

$b^2-c^2-a^2=2ca$

$c^2-a^2-b^2=2ab$

Do đó: $A=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2ab}$

$=\dfrac{a^3+b^3+c^3}{2abc}$

Mà $a+b+c=0$ nên:

$a^3+b^3+c^3=3abc$

Suy ra $A=\dfrac{3abc}{2abc}$

$A=\dfrac32$

2.

Cho $abc=2$.

Ta có:

$A=\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}$

Vì $abc=2$ nên:

$ab=\dfrac2c,\quad bc=\dfrac2a,\quad ac=\dfrac2b$

Suy ra:

$A=\dfrac{ac}{2+ac+2c}+\dfrac{ab}{2+ab+a}+\dfrac{2bc}{2+2bc+2b}$

$=\dfrac{ac}{2+ac+2c}+\dfrac{ab}{2+ab+a}+\dfrac{bc}{1+bc+b}$

Quy đồng và sử dụng $abc=2$:

$A=1$

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

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

\(=\left(b-a\right)\left(b^2-3bc+3c^2+ab-3ac+a^2\right)-\left(a-b\right)\left(1+3bc-3ab-3c^2+3ac\right)\)

\(=\left(b-a\right)\left(b^2-3bc+3c^2+ab-3ac+a^2+1+3bc-3ab-3c^2+3ac\right)\)

\(=\left(b-a\right)\left(b^2-2ab+a^2+1\right)\)

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

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

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

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

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

\(=0\)

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

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

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

\(=\left(a-b\right)^2-\left(a-b\right)^2=0\)

a + b + a + b +c + c - a - b - b + c + a - b - c -a + c + a

= a + ( - b ) + ( - c)

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

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

\(=2a-b+3c\)

k mk nha 

thank you very much