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\(P=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-c\right)\left(b-a\right)}+\frac{c^2}{\left(c-b\right)\left(c-a\right)}\)
\(=\frac{-a^2}{\left(a-b\right)\left(c-a\right)}+\frac{-b^2}{\left(b-c\right)\left(a-b\right)}+\frac{-c^2}{\left(b-c\right)\left(c-a\right)}\)
\(=\frac{\left(-a^2\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{\left(-b^2\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{\left(-c^2\right)\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{-a^2b+ca^2-b^2c+ab^2-c^2a+bc^2}{-a^2b-c^2a+ca^2-b^2c+ab^2+bc^2}=1\)
Vậy \(P=1.\)
Ta có:\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
\(\Rightarrow\frac{a}{b-c}=\frac{b}{a-c}+\frac{c}{b-a}=\frac{b^2-ab+ac-c^2}{\left(c-a\right)\left(a-b\right)}\)
\(\frac{\Leftrightarrow a}{\left(b-c\right)^2}=\frac{b^2-ab+ac-c^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(1\right)\) Nhân hai vế với \(\frac{1}{b-c}\)
Tương tự ta có:\(\frac{b}{\left(c-a\right)^2}=\frac{c^2-bc+ba-a^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(2\right);\frac{c}{\left(a-b\right)^2}=\frac{a^2-ac+bc-b^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(3\right)\)
Cộng (1),(2),(3) ta được đpcm
\(a.\) Với \(a+b+c=0\) thì \(\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{\left(-c\right).\left(-a\right).\left(-b\right)}{abc}=\frac{-abc}{abc}=-1\)
\(b.\) Công thức tổng quát: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
Ta có:
\(\frac{1}{x\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)
\(\frac{1}{\left(x+1\right)\left(x+2\right)}=\frac{1}{x+1}-\frac{1}{x+2}\)
\(\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{1}{x+2}-\frac{1}{x+3}\)
\(\frac{1}{\left(x+3\right)\left(x+4\right)}=\frac{1}{x+3}-\frac{1}{x-4}\)
\(\frac{1}{\left(x+4\right)\left(x+5\right)}=\frac{1}{x+4}-\frac{1}{x+5}\)
Do đó, suy ra được: \(A=\frac{1}{x}-\frac{1}{x+5}=\frac{x+5-x}{x\left(x+5\right)}=\frac{5}{x\left(x+5\right)}\)
\(\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{\left(c-b\right)-\left(c-a\right)}{\left(c-a\right)\left(c-b\right)}=\frac{1}{c-a}-\frac{1}{c-b}=\frac{1}{c-a}+\frac{1}{b-c}\)
Tương tự:
\(\frac{b-c}{\left(a-b\right)\left(a-c\right)}=\frac{1}{a-b}+\frac{1}{c-a};\frac{c-a}{\left(b-c\right)\left(a-b\right)}=\frac{1}{b-c}+\frac{1}{a-b}\)
Cộng lại có đpcm