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Đặt \(\left(\frac{a-b}{c},\frac{b-c}{a},\frac{c-a}{b}\right)\rightarrow\left(x,y,z\right)\)
Khi đó:\(\left(\frac{c}{a-b},\frac{a}{b-c},\frac{b}{c-a}\right)\rightarrow\left(\frac{1}{x},\frac{1}{y},\frac{1}{z}\right)\)
Ta có:
\(P\cdot Q=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3+\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)
Mặt khác:\(\frac{y+z}{x}=\left(\frac{b-c}{a}+\frac{c-a}{b}\right)\cdot\frac{c}{a-b}=\frac{b^2-bc+ac-a^2}{ab}\cdot\frac{c}{a-b}\)
\(=\frac{c\left(a-b\right)\left(c-a-b\right)}{ab\left(a-b\right)}=\frac{c\left(c-a-b\right)}{ab}=\frac{2c^2}{ab}\left(1\right)\)
Tương tự:\(\frac{x+z}{y}=\frac{2a^2}{bc}\left(2\right)\)
\(=\frac{x+y}{z}=\frac{2b^2}{ac}\left(3\right)\)
Từ ( 1 );( 2 );( 3 ) ta có:
\(P\cdot Q=3+\frac{2c^2}{ab}+\frac{2a^2}{bc}+\frac{2b^2}{ac}=3+\frac{2}{abc}\left(a^3+b^3+c^3\right)\)
Ta có:\(a+b+c=0\)
\(\Rightarrow\left(a+b\right)^3=-c^3\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
Khi đó:\(P\cdot Q=3+\frac{2}{abc}\cdot3abc=9\)
2, (trích đề thi học sinh giỏi Bến Tre-1993)
\(a^3+a^2b+ca^2+b^3+ab^2+b^2c+c^3+c^2b+c^2a=a^2\left(a+b+c\right)+b^2\left(a+b+c\right)+c^2\left(a+b+c\right)=\left(a+b+c\right)\left(a^2+b^2+c^2\right)\)
mà a+b+c=0 => (a+b+c)(a2+b2+c2)=0
=> đpcm
*bài này tui làm tắt, không hiểu ib
Vừa lm xog bị troll chứ, tuk quá
\(x-a^2x-\frac{b^2}{b^2-x^2}+a=\frac{x^2}{x^2-b^2}\)
\(\Leftrightarrow\frac{x\left(b^2-x^2\right)\left(x^2-b^2\right)}{\left(b^2-x^2\right)\left(x^2-b^2\right)}-\frac{a^2x\left(b^2-x^2\right)\left(x^2-b^2\right)}{\left(b^2-x^2\right)\left(x^2-b^2\right)}-\frac{b^2\left(x^2-b^2\right)}{\left(b^2-x^2\right)\left(x^2-b^2\right)}+\frac{a\left(b^2-x^2\right)\left(x^2-b^2\right)}{\left(b^2-x^2\right)\left(x^2-b^2\right)}=\frac{x^2\left(b^2-x^2\right)}{\left(b^2-x^2\right)\left(x^2-b^2\right)}\)
Khử mẫu :
\(\Leftrightarrow2x^3b^2-xb^4-x^5-2a^2x^3b^2+a^2xb^4+a^2x^5-b^2x^2+b^4+2ab^2x^2-ab^4-ax^4=x^2b^2-x^4\)
Tự xử nốt, lm bài này muốn phát điên mất.
c/
\(M=x^2+5y^2-4xy+2x-8y+2018\)
\(M=\left(x^2+4y^2+1-4xy+2x-4y\right)+\left(y^2-4y+4\right)+2013\)
\(M=\left(x-2y+1\right)^2+\left(y-2\right)^2+2013\ge2013\)
\(M_{min}=2013\) khi \(\left\{{}\begin{matrix}x-2y+1=0\\y-2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)
2/
\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
\(\Leftrightarrow\frac{a}{b-c}=-\frac{b}{c-a}-\frac{c}{a-b}=\frac{b}{a-c}+\frac{c}{b-a}\)
\(\Leftrightarrow\frac{a}{b-c}=\frac{b\left(b-a\right)+c\left(a-c\right)}{\left(a-c\right)\left(b-a\right)}=\frac{b^2-ab+ac-c^2}{\left(a-b\right)\left(c-a\right)}\)
\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}=\frac{b^2-ab+ac-c^2}{\left(a-b\right)\left(c-a\right)\left(b-c\right)}\)
Tương tự ta có: \(\frac{b}{\left(c-a\right)^2}=\frac{c^2+ab-bc-a^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\) ; \(\frac{c}{\left(a-b\right)^2}=\frac{a^2+bc-ac-b^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Cộng vế với vế:
\(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=\frac{b^2-ab+ac-c^2+c^2+ab-bc-a^2+a^2+bc-ca-b^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)