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a) Ta có : \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\Leftrightarrow\frac{a+b}{c}+1=\frac{b+c}{a}+1=\frac{c+a}{b}+1\)
\(\Rightarrow\frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}\)
- TH1: Nếu a + b + c = 0 \(\Rightarrow P=\frac{a+b}{b}.\frac{b+c}{c}.\frac{a+c}{a}=\frac{-c}{b}.\frac{-a}{c}.\frac{-b}{a}=\frac{-\left(abc\right)}{abc}=-1\)
- TH2 : Nếu \(a+b+c\ne0\) \(\Rightarrow a=b=c\)
\(\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
b) Đề bài sai ^^
Đặt \(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=A\)
Ta có:\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
<=> \(\left(\frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b}\right)\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)=0\)
<=> \(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(b-c\right)\left(c-a\right)}+\frac{c}{\left(b-c\right)\left(a-b\right)}+\frac{a}{\left(b-c\right)\left(c-a\right)}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)\left(c-a\right)}+\frac{a}{\left(a-b\right)\left(b-c\right)}+\frac{b}{\left(a-b\right)\left(c-a\right)}+\frac{c}{\left(a-b\right)^2}=0\)
<=> \(A+\frac{a+b}{\left(b-c\right)\left(c-a\right)}+\frac{c+a}{\left(a-b\right)\left(b-c\right)}+\frac{c+b}{\left(a-b\right)\left(c-a\right)}=0\)
<=> \(A+\frac{\left(a+b\right)\left(a-b\right)+\left(c-a\right)\left(c+a\right)+\left(c+b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
<=> \(A+\frac{a^2-b^2+c^2-a^2+b^2-c^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
<=> \(A=0\)
=> ....
Ta có : \(\frac{a-\left(c-b\right)}{b-c}+\frac{b-\left(a-c\right)}{c-a}+\frac{c-\left(b-a\right)}{a-b}=3\)
\(\Leftrightarrow\frac{a+\left(b-c\right)}{b-c}-1+\frac{b+\left(c-a\right)}{c-a}-1+\frac{c+\left(a-b\right)}{a-b}-1=0\)
\(\Leftrightarrow\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
\(\Rightarrow\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\left(\frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b}\right)=0\)
\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(a-c\right)^2}+\frac{c}{\left(a-b\right)^2}+\frac{a+b}{\left(b-c\right)\left(c-a\right)}+\frac{a+c}{\left(b-c\right)\left(a-b\right)}+\frac{b+c}{\left(c-a\right)\left(a-b\right)}=0\)
\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}+\frac{a^2-b^2+c^2-a^2+b^2-c^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)
Từ gt ta có : \(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)0
Từ đó suy ra điều phải chứng minh
Ta có
\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
\(\Rightarrow\frac{a}{b-c}=\frac{b}{a-c}+\frac{c}{b-a}=\frac{b^2-ab+ac-c^2}{\left(a-c\right)\left(b-a\right)}\)
\(\Rightarrow\frac{a}{\left(b-c\right)^2}=\frac{b^2-ab+ac-c^2}{\left(a-c\right)\left(b-a\right)\left(b-c\right)}\)
Tương tự
\(\frac{b}{\left(c-a\right)^2}=\frac{c^2-bc+ab-a^2}{\left(a-c\right)\left(b-a\right)\left(b-c\right)}\)
\(\frac{c}{\left(a-b\right)^2}=\frac{a^2-ac+bc-b^2}{\left(a-c\right)\left(b-a\right)\left(b-c\right)}\)
\(\Rightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=\frac{b^2-ab+ac-c^2+c^2-bc+ab-a^2+a^2-ac+bc-b^2}{\left(a-c\right)\left(b-a\right)\left(a-b\right)}\)
=0 ( ĐPCM)