Nâng cao và phát triển toán 8 tập 1 bài 153
\(\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\frac{1}{b-c}=0\Rightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)\left(b-c\right)}+\frac{c}{\left(a-b\right)\left(b-c\right)}=0\)(1)
\(\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\frac{1}{c-a}=0\Rightarrow\frac{a}{\left(b-c\right)\left(c-a\right)}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)\left(c-a\right)}=0\)(2)
\(\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\frac{1}{a-b}=0\Rightarrow\frac{a}{\left(b-c\right)\left(a-b\right)}+\frac{b}{\left(c-a\right)\left(a-b\right)}+\frac{c}{\left(a-b\right)^2}=0\)(3)
Cộng (1);(2);(3) ta có: \(M+\frac{b\left(a-b\right)+c\left(c-a\right)+a\left(a-b\right)+c\left(b-c\right)+a\left(c-a\right)+b\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\Rightarrow\)\(M+\frac{ab-b^2+c^2-ac+a^2-ab+bc-c^2+ac-a^2+b^2-bc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
\(\Rightarrow M+0=0\Rightarrow M=0\)
tick cho mk nka ^^