\(=\dfrac{b\left(b-c\right)-a\left(a-c\right)}{ab\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

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

\(=\dfrac{-\left(a-b\right)\left(a+b\right)+c\left(a-b\right)}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\dfrac{-a-b+c}{ab\left(a-c\right)\left(b-c\right)}\)

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

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

\(=\dfrac{\left(b-a\right)\left(b+a\right)-c\left(b-a\right)}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\dfrac{-\left(b+a-c\right)}{ab\left(a-c\right)\left(b-c\right)}\)

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

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

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

\(=\dfrac{-\left(a-b\right)\left(a+b\right)+c\left(a-b\right)}{ab\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\dfrac{\left(a-b\right)\left(-a-b+c\right)}{ab\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=-\dfrac{a+b-c}{ab\left(b-c\right)\left(a-c\right)}\)

thiếu đề bài òi bạn ko làm đc đâu

Mik giải ra rồi!

giải kiểu gì ??????

😥 😥 😥

Ta có: \(A=\left(a+b-c\right)+\left(a-b\right)-\left(a-b\right)-\left(a-b-c\right).\)

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

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

\(\Rightarrow A=3.\left(a+b\right)\)

Rút gọn biểu thức :

 A=(a+b−c)+(a−b)−(a−b)−(a−b−c)

Ta có: A=(a+b−c)+(a−b)−(a−b)−(a−b−c).

⇒A=a+b−c+a+b+a+b+a+b+c

⇒A=a+b+a+b+a+b+a+b

⇒A=3.(a+b)

 nhé !

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

\(=2b.\)

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

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

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

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

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

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

C=3a²-ab+ac+3ab-b²+bc+3ac-bc+c²

C=3a²-b²+c²+2ab+4ac

C=3a²-b²+c²+2a(b+2c)

mk thấy có một số cái giống đó bn nhóm lại thành một được không???

mk cx chả bít!! 876869789780

ta có : a+b+c=0=>a+b=-c ; b+c=-a ; a+c=-b 

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

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

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

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

M=\(\frac{2ab}{-2ab}+\frac{2bc}{-2bc}+\frac{2ca}{-2ca}\)

M=-1-1-1=-3

Vậy với a+b+c=0 thì M=-3