b ≠ c,a+b≠c, c²+2ab-2ac-2bc=0 rút gọn (a ² +(a-c) ²)/(b ²+(b-c) ²)

Giải thích các bước giải:

Ta có:

$\dfrac{a^2+(a-c)^2}{b^2+(b-c)^2}$

$=\dfrac{a^2+c^2+2ab-2ac-2bc+(a-c)^2}{b^2+c^2+2ab-2ac-2bc+(b-c)^2}$

$=\dfrac{(a-c)^2+2b(a-c)+(a-c)^2}{(b-c)^2+2a(b-c)+(b-c)^2}$

$=\dfrac{(a-c)(a-c+2b+a-c)}{(b-c)(b-c+2a+b-c)}$

$=\dfrac{(a-c)(2a+2b-2c)}{(b-c)(2a+2b-2c)}$

$=\dfrac{a-c}{b-c}$