Question

Cho \(c^2+2ab-2bc-2ca=0\)    \(b\ne c;a+b\ne c\)

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

Answer 1

theo bài ra ta có: \(c^2+2ab-2bc-2ca=0.\)

\(\Rightarrow2\left(c^2+ab-bc-ca\right)=c^2\)

\(\Rightarrow2\left(a-c\right)\left(b-c\right)=c^2\)

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

                                                                               \(=\frac{\left(a-c\right)\left(a+b-c\right)}{\left(b-c\right)\left(b+a-c\right)}=\frac{a-c}{b-c}\) => đpcm