Question

Cho c² + 2(ab - ac - bc) = 0; b khác c, a + b khác c.

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

Answer 1

Ta có:

Ta có:

(b-c)(a-c)(a+b)+(a-c)(a+b)(c-b)=0

suy ra

(b-c)(a^2+ab-ac-bc)+(a-c)(ac+bc-ab-b^2)=0

Ta có: c^2 +2(ab-ac-bc)=0

suy ra:

ab-ac-bc=-c^2+ab+bc

c^2+ab-bc-ac=ac+bc-ab

Vậy (b-c)(a^2-c^2+ab+bc+ac)+(a-c)(ab-bc-ac+c^2-b^2)=0

suy ra:(b-c)[a^2-(a-c)(b-c)]+(a-c)[(a-c)(b-c)-b^2]=0

suy ra a^2(b-c)-(a-c)(b-c)^2+(a-c)^2.(b-c)-(a-c).b^2=0

Suy ra a^2(b-c)+(a-c)^2.(b-c)=(a-c).b^2+(a-c)(b-c)^2

suy ra:(b-c)(a^2+(a-c)^2)=(a-c)(b^2+(b-c)^2)

suy ra a-c/b-c=a^2+(a-c)^2/b^2+(b-c)^2(đpcm)