hàiiii chán quá

ĐK : \(a\ne b\ne c\)

\(\dfrac{a^3+b^3+c^3-3abc}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)

\(=\dfrac{\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)

\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-bc-ca\right)-3ab\left(a+b+c\right)}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)

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

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

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

\(=\dfrac{\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}{2\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}\)

\(=\dfrac{a+b+c}{2}\)

Lời giải:

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

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

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

=\(2a^3+6a\left(b-c\right)^2\) +\(\left(b+c\right)^3-3\left(b+c\right)^2\cdot a+3\left(b+c\right)\cdot a^2-a^3\) -\(a^3-3a^2\left(b+c\right)-3a\left(b+c\right)^2-\left(b+c\right)^3\)

\(=2a^3+6a\left(b-c\right)^2-6a\left(b+c\right)^2-2a^3\)

\(=6a\left(b-c\right)^2-6a\left(b+c\right)^2\)

=6a[(b-c-b-c)(b-c+b+c)]

=6a(-2b)*2c

=-24abc