Question

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

Answer 1

cầm cm cái này trước đã

\(\dfrac{1}{a-b}-\dfrac{1}{a-c}=\dfrac{a-c}{a-b}+\dfrac{a-b}{a-c}=\dfrac{b-c}{\left(a-b\right)\left(a-c\right)}\)

áp dụng vào bài

\(=>\left\{{}\begin{matrix}\dfrac{b-c}{\left(a-b\right)\left(a-c\right)}=\dfrac{1}{a-b}-\dfrac{1}{a-c}\\\dfrac{c-a}{\left(b-c\right)\left(b-a\right)}=\dfrac{1}{b-c}-\dfrac{1}{b-a}\\\dfrac{a-b}{\left(c-a\right)\left(c-b\right)}=\dfrac{1}{c-a}-\dfrac{1}{c-b}\end{matrix}\right.\)

thay vào đề,

=> đpcm

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