b ≠ c,a+b≠c, c²+2ab-2ac-2bc=0 rút gọn (a ² +(a-c) ²)/(b ²+(b-c) ²) 16/08/2021 Bởi Harper 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: $\dfrac{a^2+(a-c)^2}{b^2+(b-c)^2}$ $=\dfrac{a^2+c^2+2ab-2ac-2bc+(a-c)^2}{b^2+c^2+2ab-2ac-2bc+(b-c)^2}$ $=\dfrac{(a-c)^2+2b(a-c)+(a-c)^2}{(b-c)^2+2a(b-c)+(b-c)^2}$ $=\dfrac{(a-c)(a-c+2b+a-c)}{(b-c)(b-c+2a+b-c)}$ $=\dfrac{(a-c)(2a+2b-2c)}{(b-c)(2a+2b-2c)}$ $=\dfrac{a-c}{b-c}$ Bình luận
Giải thích các bước giải:
Ta có:
$\dfrac{a^2+(a-c)^2}{b^2+(b-c)^2}$
$=\dfrac{a^2+c^2+2ab-2ac-2bc+(a-c)^2}{b^2+c^2+2ab-2ac-2bc+(b-c)^2}$
$=\dfrac{(a-c)^2+2b(a-c)+(a-c)^2}{(b-c)^2+2a(b-c)+(b-c)^2}$
$=\dfrac{(a-c)(a-c+2b+a-c)}{(b-c)(b-c+2a+b-c)}$
$=\dfrac{(a-c)(2a+2b-2c)}{(b-c)(2a+2b-2c)}$
$=\dfrac{a-c}{b-c}$