Question

Cho \(c^2+2\left(ab-ac-bc\right)=0;b\ne c;a+b\ne c\)thì 

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

Answer 1

thế cuối cùng đề bài là gì'.'???????

Answer 2

Đề rõ vậy còn gì Chứng minh

Answer 3

Ta có: \(c^2+2\left(ab-ac-bc\right)=0\)

        \(\Rightarrow c^2=-2\left(ab-ac-bc\right)\)

Thay vào 

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

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

=> ...