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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:\(P\cdot Q=3+\frac{2}{abc}\cdot3abc=9\)
từ đề bài \(\Rightarrow\frac{a}{b-c}=-\frac{b}{c-a}-\frac{c}{a-b}=\frac{-b\left(a-b\right)-c\left(c-a\right)}{\left(a-b\right)\left(c-a\right)}=\frac{-ab+b^2-c^2+ac}{\left(a-b\right)\left(c-a\right)}\)
\(\Rightarrow\frac{a}{\left(b-c\right)^2}=\frac{-ab+b^2-c^2+ac}{\left(a-b\right)\left(c-a\right)\left(b-c\right)}\)
Tương tự : \(\hept{\begin{cases}\frac{b}{\left(c-a\right)^2}=\frac{-cb+c^2-a^2+ab}{\left(a-b\right)\left(c-a\right)\left(b-c\right)}\\\frac{c}{\left(a-b\right)^2}=\frac{-ac+a^2-b^2+bc}{\left(a-b\right)\left(c-a\right)\left(b-c\right)}\end{cases}}\)
Cộng vế với vế ta được : \(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c^2}{\left(a-b\right)^2}\)
\(=\frac{-ab+b^2-c^2+ac-bc+c^2-a^2+ab-ac+a^2-b^2+bc}{\left(a-b\right)\left(c-a\right)\left(b-c\right)}=0\)(đpcm)
Đặt \(P=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\) ; \(Q=\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\)
Ta có : \(P=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}=\frac{ab\left(a-b\right)+bc\left(b-c\right)+ac\left(c-a\right)}{abc}\)
Xét tử số của P : \(ab\left(a-b\right)+bc\left(b-c\right)+ac\left(c-a\right)=ab\left[-\left(b-c\right)-\left(c-a\right)\right]+bc\left(b-c\right)+ac\left(c-a\right)\)
\(=-ab\left(b-c\right)-ab\left(c-a\right)+bc\left(b-c\right)+ac\left(c-a\right)\)
\(=b\left(b-c\right)\left(c-a\right)+a\left(c-a\right)\left(c-b\right)=\left(b-c\right)\left(c-a\right)\left(b-a\right)\)
\(\Rightarrow P=\frac{\left(b-c\right)\left(c-a\right)\left(b-a\right)}{abc}\)
Lại có : \(Q=\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\). Đặt \(a-b=x\); \(b-c=y\); \(c-a=z\)
Suy ra được : \(\hept{\begin{cases}x-y=a-b-b+c=a+c-2b=-3b\\y-z=b-c-c+a=a+b-2c=-3c\\z-x=c-a+b-a=b+c-2a=-3a\end{cases}\Rightarrow\hept{\begin{cases}b=-\frac{\left(x-y\right)}{3}\\c=-\frac{\left(y-z\right)}{3}\\a=-\frac{\left(z-x\right)}{3}\end{cases}}}\)
Ta có : \(Q=\frac{-\left(\frac{y-z}{3}\right)}{x}+\frac{-\left(\frac{z-x}{3}\right)}{y}+\frac{-\left(\frac{x-y}{3}\right)}{z}=-\frac{1}{3}.\left(\frac{y-z}{x}+\frac{z-x}{y}+\frac{x-y}{z}\right)\)
\(=-\frac{1}{3}\left(\frac{yz\left(y-z\right)+xz\left(z-x\right)+yx\left(x-y\right)}{xyz}\right)\)
Đến đây rút gọn tương tự với P được: \(Q=\frac{\left(x-z\right)\left(x-y\right)\left(z-y\right)}{3xyz}=\frac{\left(3a\right).\left(-3b\right).\left(3c\right)}{3\left(a-b\right)\left(b-c\right)\left(c-a\right)}\Rightarrow Q=\frac{-9abc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Vậy : \(PQ=\frac{\left(b-c\right)\left(c-a\right)\left(b-a\right)}{abc}.\frac{-9abc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=9\)
Vậy ta có điều phải chứng minh.
\(\)
Ta có: \(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
\(\Rightarrow\frac{a}{b-c}=-\frac{b}{c-a}-\frac{c}{a-b}\)
\(\Rightarrow\frac{a}{b-c}=\frac{-b\left(a-b\right)-c\left(c-a\right)}{\left(c-a\right)\left(a-b\right)}\)
\(\Rightarrow\frac{a}{b-c}=\frac{-ab+b^2-c^2+ac}{\left(c-a\right)\left(a-b\right)}\)
\(\Rightarrow\frac{a}{\left(b-c\right)^2}=\frac{-ab+b^2-c^2+ac}{\left(c-a\right)\left(a-b\right)\left(b-c\right)}\)
Tương tự ta có: \(\frac{b}{\left(c-a\right)^2}=\frac{-bc+c^2-a^2+ab}{\left(c-a\right)\left(a-b\right)\left(b-c\right)}\)
\(\frac{c}{\left(a-b\right)^2}=\frac{-ca+a^2-b^2+bc}{\left(c-a\right)\left(a-b\right)\left(b-c\right)}\)
Cộng các đẳng thức trên ta được:
\(\frac{a}{\left(b-c\right)^2}\)\(+\frac{b}{\left(c-a\right)^2}\)\(+\frac{c}{\left(a-b\right)^2}=\)\(\frac{-ab+b^2-c^2+ac-bc+c^2-a^2+ba-ca+a^2-b^2+bc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{0}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
Vậy \(\frac{a}{\left(b-c\right)^2}\)\(+\frac{b}{\left(c-a\right)^2}\)\(+\frac{c}{\left(a-b\right)^2}=\)0 (đpcm)
Câu hỏi của Jungkookie - Toán lớp 7 - Học toán với OnlineMath
Ta có ; \(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)(1)
Nhân cả hai vế của (1) với \(\frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b}\)được :
\(\Leftrightarrow\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\left(\frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b}\right)=0\)
\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{a}{\left(b-c\right)\left(c-a\right)}+\frac{a}{\left(a-b\right)\left(b-c\right)}+\frac{b}{\left(c-a\right)\left(b-c\right)}+\frac{b}{\left(c-a\right)^2}+\frac{b}{\left(c-a\right)\left(a-b\right)}+\frac{c}{\left(a-b\right)\left(b-c\right)}+\frac{c}{\left(a-b\right)\left(c-a\right)}+\frac{c}{\left(a-b\right)^2}=0\)
\(\Leftrightarrow\left[\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}\right]+\frac{a+b}{\left(b-c\right)\left(c-a\right)}+\frac{b+c}{\left(a-b\right)\left(c-a\right)}+\frac{c+a}{\left(a-b\right)\left(b-c\right)}=0\)\(\Leftrightarrow\left[\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}\right]+\frac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)
Vậy ta có điều phải chứng minh.