Đặt x = a+b , y = b+c , z = c+a

Thì biểu thức trên trở thành \(x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3xy-3xyz\)

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

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

Từ đó thay a,b,c vào rồi rút gọn :)

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

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

giúp với please

\(-\frac{a-c}{a+c}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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