Question

Cho 3 số a,b,c tm: c² + 2.(ab-bc-ca)=0 , b \(\ne\) c , a+b \(\ne\) c. CM

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

Answer 1

\(\left\{{}\begin{matrix}c^2-2ca+a^2+2ab-2bc=a^2\\c^2-2bc+b^2+2ab-2ac=b^2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(a-c\right)^2+2b\left(a-c\right)=a^2\\\left(b-c\right)^2+2a\left(b-c\right)=b^2\end{matrix}\right.\)

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

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