Question

Rút gọn biểu thức

M=1 trên (a-b)(a-c)+1trên (b-c)(b-a)+1trên(c-a)(c-b)

Answer 1

ĐKXĐ: \(a\ne b\ne c\)

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

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

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

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

\(=0\)

Answer 2

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

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

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

\(M=\frac{1\left(b-c\right)-1\left(a-c\right)+1\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=\frac{b-c-a+c+a-b}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=\frac{0}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=0\)