Question

[Macoto Kino]

cm

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

Answer 1

[( էҽąണ ๖ۣۜTαм ๖ۣۜGĭᢠ๖ۣۜQυỷ )____...]

đây là toán mà bn

Answer 2

[Nguyễn Linh Chi]

Đặt: b + c - a = x; a + b - c = y; a + c - b = z

khi đó: x + y + z = a + b + c 

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

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

Ta cần chứng minh: 

\(a\left(b-c\right)\left(b+c-a\right)^2+c\left(a-b\right)\left(a+b-c\right)^2=b\left(a-c\right)\left(a+c-b\right)^2\)(1)

<=> \(a\left(b-c\right)\left(b+c-a\right)^2+c\left(a-b\right)\left(a+b-c\right)^2+b\left(c-a\right)\left(a+c-b\right)^2=0\)

Hay mình cần chứng minh: 

\(\frac{y+z}{2}.\frac{y-z}{2}.x^2+\frac{z+x}{2}.\frac{z-x}{2}.y^2+\frac{x+y}{2}.\frac{x-y}{2}.z^2=0\)

<=> \(\left(y^2-z^2\right)x^2+\left(z^2-x^2\right)y^2+\left(x^2-y^2\right)z^2=0\)

<=> \(0=0\)luôn đúng

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