{Toán 8} Rút gọn!

T

transformers123

Đề hình như sai rồi, phải là:
$\dfrac{1}{(b-c)(a^2+ac-b^2-bc)}+\dfrac{1}{(c-a)(b^2+ab-c^2-ac)}+\dfrac{1}{(a-b)(c^2+bc-a^2-ab)}$
$=\dfrac{1}{(b-c)(a-b)(a+b+c)}+\dfrac{1}{(c-a)(b-c)(a+b+c)}+\dfrac{1}{(a-b)(c-a)(a+b+c)}$
$=\dfrac{c-a+a-b+b-c}{(a-b)(b-c)(c-a)(a+b+c)}$
$=\dfrac{0}{(a-b)(b-c)(c-a)(a+b+c)}$
$=0$
 
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