Question

Cho a, b,c là các số thực đôi một khác nhau. Hãy tính giá trị của biểu thức:

\(M=\dfrac{a+b}{a-b}.\dfrac{b+c}{b-c}+\dfrac{b+c}{b-c}.\dfrac{c+a}{c-a}+\dfrac{c+a}{c-a}.\dfrac{a+b}{a-b}\)

Answer 1

\(M=\dfrac{\left(a+b\right)\left(b+c\right)\left(c-a\right)+\left(b+c\right)\left(c+a\right)\left(a-b\right)+\left(c+a\right)\left(a+b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\dfrac{\left(a+b\right)\left(\left(b+c\right)\left(c-a\right)+\left(c+a\right)\left(b-c\right)\right)+\left(b+c\right)\left(c+a\right)\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\dfrac{\left(a+b\right)\left(2bc-2ac\right)+\left(b+c\right)\left(c+a\right)\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)} =\dfrac{-2c\left(a+b\right)\left(a-b\right)+\left(a-b\right)\left(b+c\right)\left(c+a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\dfrac{\left(a-b\right)\left(-2ac-2bc+bc+ab+c^2+ac\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\dfrac{\left(a-b\right)\left(ab-bc-ac+c^2\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\dfrac{-\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=-1\)