Ta có : \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}\)= \(\dfrac{1}{a-b}-\dfrac{1}{b-c}\)
\(\dfrac{1}{\left(b-c\right)\left(c-a\right)}=\dfrac{1}{b-c}-\dfrac{1}{c-a}\)
\(\dfrac{1}{\left(c-a\right)\left(a-b\right)}=\dfrac{1}{c-a}-\dfrac{1}{a-b}\)
=> \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)} =\dfrac{1}{a-b}-\dfrac{1}{b-c}+\dfrac{1}{b-c}-\dfrac{1}{c-a}+\dfrac{1}{c-a}-\dfrac{1}{a-b}\)= \(\dfrac{1}{a-b}-\dfrac{1}{a-b}\) = 0
P/S: Mình ko biết có đúng ko nên kiểm tra lại giúp mình nha :3
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