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

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

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

\(=4a-4b\)

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

\(\Leftrightarrow\frac{bc+ac+ba}{abc}=0\Leftrightarrow bc+ac+ba=0\Leftrightarrow c.\left(a+b\right)=-ba\Leftrightarrow a+b=\frac{-ab}{c}\)

\(b+c=-\frac{bc}{a},a+c=\frac{-ac}{b}\)

thay vô là đc :") lazzy~~

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 $abc=2$:

$A=1$

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

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

\(=\frac{b-c-a+c+a-b}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=0\)

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

= a(b - c) + 1

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

= b(c - a) + 1

(c - a)(c - b)

= c(a - b)

học tốt!

Bài 1 : 

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

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

Ta có b = -1 ta được : \(2b=2\left(-1\right)=-2\)

Vậy \(A=-2\)

\(B=\left(-2a+3b-4c\right)-\left(-2a-3b-4c\right)=-2a+3b-4c+2a+3b+4c\)

\(=6b\)

Ta có : b = -1 khi đó: \(B=6b=6\left(-1\right)=-6\)

Vậy B = -6