e Bon đz:
ĐKXĐ: \(a\ne b\ne c\) =))
\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
\(\Rightarrow\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)=0\)
Nhân hết ra và tự làm tiếp nhé~
Ta có : \(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
\(\Leftrightarrow\frac{a}{b-c}=\frac{-b}{c-a}-\frac{c}{a-b}\)
\(\Leftrightarrow\frac{a}{b-c}=-\left(\frac{b\left(a-b\right)}{\left(c-a\right)\left(a-b\right)}+\frac{c\left(a-b\right)}{\left(c-a\right)\left(a-b\right)}\right)\)
\(\Leftrightarrow\frac{a}{b-c}=-\left(\frac{ab-b^2+c^2-ac}{\left(c-a\right)\left(a-b\right)}\right)\left(1\right)\)
Hoàn toàn tương tự với hai phân thức còn lại ta có :
\(\frac{b}{c-a}=-\left(\frac{a^2-ab+bc-c^2}{\left(b-c\right)\left(a-b\right)}\right)\left(2\right)\)
\(\frac{c}{a-b}=-\left(\frac{b^2-bc+ac-a^2}{\left(c-a\right)\left(b-c\right)}\right)\left(3\right)\)
Mặt khác :
\(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}\)
\(=\frac{a}{b-c}\cdot\frac{1}{b-c}+\frac{b}{c-a}\cdot\frac{1}{c-a}+\frac{c}{a-b}\cdot\frac{1}{a-b}\)
Thay (1), (2) và (3) vào phân thức trên ta được :
\(-\left(\frac{ab-b^2+c^2-ac}{\left(c-a\right)\left(a-b\right)}\right)\cdot\frac{1}{b-c}-\left(\frac{a^2-ab+bc-c^2}{\left(b-c\right)\left(a-b\right)}\right)\cdot\frac{1}{c-a}-\left(\frac{b^2-bc+ac-a^2}{\left(c-a\right)\left(b-c\right)}\right)\cdot\frac{1}{a-b}\)
\(=\frac{-ab+b^2-c^2+ac}{\left(c-a\right)\left(a-b\right)\left(b-c\right)}+\frac{-a^2+ab-bc+c^2}{\left(b-c\right)\left(a-b\right)\left(c-a\right)}+\frac{-b^2+bc-ac+a^2}{\left(c-a\right)\left(b-c\right)\left(a-b\right)}\)
\(=\frac{-ab+b^2-c^2+ac-a^2+ab-bc+c^2-b^2+bc-ac+a^2}{\left(c-a\right)\left(a-b\right)\left(b-c\right)}\)
\(=\frac{0}{\left(c-a\right)\left(a-b\right)\left(b-c\right)}\)
\(=0\left(đpcm\right)\)
@kudo shinichi cách của anh thì đc nhưng làm thế nào để tìm được cái thừa số bên trái ấy ?
Hihi cách em logic hơn nhé :D
Cảm ơn anh :))
Sao gõ nhanh zậy
bài lm thấy hợp lí đấy đúng òi đó
Thưởng kudo shinichi 3sp vì giúp bạn làm bài khó nhé :))
Có: \(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
\(\Rightarrow\left(\frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b}\right)\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)=0\)
\(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(b-c\right)\left(c-a\right)}+\frac{c}{\left(b-c\right)\left(a-b\right)}+\frac{b}{\left(c-a\right)^2}+\frac{a}{\left(b-c\right)\left(c-a\right)}+\frac{c}{\left(a-b\right)\left(c-a\right)}+\)
\(\frac{c}{\left(a-b\right)^2}+\frac{a}{\left(b-c\right)\left(a-b\right)}+\frac{b}{\left(a-b\right)\left(c-a\right)}=0\)
\(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}+\frac{\left(a-b\right)b+c\left(c-a\right)+a\left(a-b\right)+c\left(b-c\right)+a\left(c-a\right)+b\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(=0\)
\(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}+\frac{ab-b^2+c^2-ac+a^2-ab+bc-c^2+ac-a^2+b^2-bc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
\(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}+\frac{0}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
\(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)
đpcm
Hinh nhu cach cua a nhanh hon e Bon nhi? =))
Eh kudo shinichi phai dung dau
Khi va chi khi( tuong duong) chu
