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ta có : a+b+c=0=>a+b=-c ; b+c=-a ; a+c=-b
ta có: M= \(\frac{2ab}{a^2+\left(b+c\right)\left(b-c\right)}+\frac{2bc}{b^2+\left(c+a\right)\left(c-a\right)}+\frac{2ca}{c^2+\left(a+b\right)\left(a-b\right)}\)
M=\(\frac{2ab}{a^2-a\left(b-c\right)}+\frac{2bc}{b^2-b\left(c-a\right)}+\frac{2ca}{c^2-c\left(a-b\right)}\)
M=\(\frac{2ab}{a\left(a-b+c\right)}+\frac{2bc}{b\left(b-c+a\right)}+\frac{2ca}{c\left(c-a+b\right)}\)
M=\(\frac{2ab}{-ab+\left(a+c\right)}+\frac{2bc}{-bc+\left(a+b\right)}+\frac{2ac}{-ac+\left(b+c\right)}\)
M=\(\frac{2ab}{-2ab}+\frac{2bc}{-2bc}+\frac{2ca}{-2ca}\)
M=-1-1-1=-3
Vậy với a+b+c=0 thì M=-3
Vì \(a+b+c=0\) \(\Rightarrow\) \(c=-a-b\)
Gọi \(M=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\) , ta có:
\(M.\frac{c}{a-b}=1+\frac{c}{a-b}\left(\frac{b-c}{a}+\frac{c-a}{b}\right)=1+\frac{c}{a-b}.\frac{\left(b^2-bc+ac-a^2\right)}{ab}=1+\frac{c}{a-b}.\frac{\left(a-b\right)\left(c-a-b\right)}{ab}=1+\frac{2c^2}{ab}=1+\frac{2c^3}{abc}\)
Tương tự, \(M.\frac{a}{b-c}=1+\frac{2a^3}{abc};\) \(M.\frac{b}{c-a}=1+\frac{2b^3}{abc}\)
Mặt khác, ta cũng có: từ \(a+b+c=0\), suy ra \(a^3+b^3+c^3=3abc\)
Vậy, \(B=3+\frac{2\left(a^3+b^3+c^3\right)}{abc}=3+\frac{2.3abc}{abc}=3+6=9\) (vì \(a,b,c\ne0\) nên \(abc\ne0\) )
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)
1) \(M=a^2b^2c^2\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)
Em chú ý bài toán sau nhé: Nếu a+b+c=0 <=> \(a^3+b^3+c^3=3abc\)
CM: có:a+b=-c <=> \(\left(a+b\right)^3=-c^3\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\)
Chú ý: a+b=-c nên \(a^3+b^3+c^3=3abc\)
Do \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
Thay vào biểu thwusc M ta được M=3abc (ĐPCM)
2, em có thể tham khảo trong sách Nâng cao phát triển toán 8 nhé, anh nhớ không nhầm thì bài này trong đó
Nếu không thấy thì em có thể quy đồng lên mà rút gọn