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

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

\(=\frac{b-c-a+c+a-b}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=0\)

(a - b)(a - c) + 1

= a(b - c) + 1

(b - c)(b - a) + 1

= b(c - a) + 1

(c - a)(c - b)

= c(a - b)

học tốt!

\(ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\)

\(=ab\left(a-b\right)+bc\left(b-c\right)-ca\left(a-c\right)\)

\(=ab\left(a-b\right)+bc\left(b-c\right)-ca\left(a-b+b-c\right)\)

\(=ab\left(a-b\right)+bc\left(b-c\right)-ca\left(a-b\right)-ca\left(b-c\right)\)

\(=\left(a-b\right)\left(ab-ca\right)+\left(b-c\right)\left(bc-ca\right)\)

\(=\left(a-b\right)a\left(b-c\right)+\left(b-c\right)c\left(b-a\right)\)

\(=\left(a-b\right)a\left(b-c\right)-\left(b-c\right)c\left(a-b\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)

mình làm vội, có chỗ nào sai bạn thông cảm nha

\(A=\frac{a+b}{a^3+b^3}=\frac{a+b}{\left(a+b\right)\left(a^2-ab+b^2\right)}=\frac{1}{a^2-ab+b^2}\)

\(C=\frac{2ab-b}{8a^3-1}=\frac{b\left(2a-1\right)}{\left(2a-1\right)\left(4a^2+2a+1\right)}=\frac{b}{4a^2+2a+1}\)

Câu b xem lại đề đi nhé

cảm ơn bạn nhé

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

\(=\dfrac{b^2-bc-a^2+ac}{ab\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

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

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

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

\(=\dfrac{b^2-cb-a^2+ac}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

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

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

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

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

\(=\dfrac{b\left(b-c\right)-a\left(a-c\right)}{ab\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\dfrac{b^2-bc-a^2+ac}{ab\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

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

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

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