phương trình

B

becon_chibichibi

K

kenhaui

Câu $a$ :$\frac{x-ab}{a+b}$ + $\frac{x-ca}{c+a}$ +$\frac{x- bc}{b+c}$ = $a+b+c$
ĐKXĐ : $a+b ; c+a ,b+c $ khác $0$
\Rightarrow$(\frac{x-ab}{a+b} $- $c)$ +$(\frac{x-ca}{c+a} $- $b)$ +$(\frac{x-bc}{b+c} $- $a)$ = $0$
\Rightarrow $\frac{a-ab-bc-ca}{a+b}$ + $\frac{x-ab-bc-ca}{c+a}$ +$\frac{a-ab-bc-ca}{b+c}$ = $0$
\Rightarrow $(x-ab-ca-bc)$ $(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a})$=0
....
 
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