Question

rút gọn

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

Answer 1

mới đọc đề đã thấy rắc rối rồi

Answer 2

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

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

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

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

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