Question

[Ai Ai Ai]

Bài 1: Chứng minh rằng 

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

Answer 1

[Lương Ngọc Anh]

Đặt a+b-c=x;b+c-a=y;c+a-b=z

=>\(a=\frac{x+z}{2};b=\frac{x+y}{2};c=\frac{y+z}{2}\)

Cần Cm: a(b-c)(b+c-a)²+c(a-b)(a+b-c)²-b(a-c)(a+c-b)²=0

=> \(\frac{x+z}{2}\left(\frac{x+y}{2}-\frac{y+z}{2}\right)^2\cdot y^2+\frac{y+z}{2}\left(\frac{x+z}{2}-\frac{x+y}{2}\right)\cdot x^2-\frac{y+x}{2}\cdot\left(\frac{z+y}{2}-\frac{z+x}{2}\right)^2\cdot z^2=0\)

=>\(\frac{1}{4}\left(x^2-z^2\right)\cdot y^2+\frac{1}{4}\cdot\left(z^2-y^2\right)\cdot x^2-\frac{1}{4}\left(x^2-y^2\right)\cdot z^2=0\)(luôn đúng)

=> đpcm