Giả sử a,b,c là 3 số khác nhau và \(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
CMR
\(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)
K
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CB
2 tháng 5 2017
\(\frac{a}{b-c}=-\frac{b}{c-a}-\frac{c}{a-b}=-\frac{b\left(a-b\right)+c\left(c-a\right)}{\left(c-a\right)\left(a-b\right)}\Rightarrow\frac{a}{\left(b-c\right)^2}=-\frac{b\left(a-b\right)+c\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-c\right)}\)
sau đó chứng minh tương tự và cộng theo từng vế thôi
KK
2 tháng 1 2017
Ta có
\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
\(\Rightarrow\left\{\begin{matrix}\frac{a}{b-c}=-\frac{b}{c-a}-\frac{c}{a-b}\\\frac{b}{c-a}=-\frac{a}{b-c}-\frac{c}{a-b}\\\frac{c}{a-b}=-\frac{a}{b-c}-\frac{b}{c-a}\end{matrix}\right.\) (1)
Mà
\(\left\{\begin{matrix}\frac{a}{\left(b-c\right)^2}=\frac{a}{b-c}.\frac{1}{b-c}\\\frac{b}{\left(c-a\right)^2}=\frac{b}{c-a}.\frac{1}{c-a}\\\frac{c}{\left(a-b\right)^2}=\frac{c}{a-b}.\frac{1}{a-b}\end{matrix}\right.\)
Ta có : \(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)
\(\Rightarrow\frac{a}{b-c}.\frac{1}{b-c}+\frac{b}{c-a}.\frac{1}{c-a}+\frac{c}{a-b}.\frac{1}{a-b}=0\)
Thay điều (1) vào biểu thức ta có :
\(\frac{a}{b-c}.\frac{1}{b-c}+\frac{b}{c-a}.\frac{1}{c-a}+\frac{c}{a-b}.\frac{1}{a-b}=0\)
\(\Rightarrow\left(-\frac{b}{c-a}-\frac{c}{a-b}\right).\frac{1}{b-c}+\left(-\frac{a}{b-c}-\frac{c}{a-b}\right).\frac{1}{c-a}+\left(-\frac{a}{b-c}-\frac{b}{c-a}\right).\frac{1}{a-b}=0\)
\(\Rightarrow-\frac{b}{\left(c-a\right)\left(b-c\right)}-\frac{c}{\left(a-b\right)\left(b-c\right)}-\frac{a}{\left(b-c\right)\left(c-a\right)}-\frac{c}{\left(a-b\right)\left(c-a\right)}-\frac{a}{\left(b-c\right)\left(a-b\right)}-\frac{b}{\left(c-a\right)\left(a-b\right)}=0\)
\(\Rightarrow-\frac{b}{\left(c-a\right)\left(b-c\right)}-\frac{a}{\left(c-a\right)\left(b-c\right)}-\frac{c}{\left(a-b\right)\left(b-c\right)}-\frac{a}{\left(a-b\right)\left(b-c\right)}-\frac{c}{\left(c-a\right)\left(a-b\right)}-\frac{b}{\left(c-a\right)\left(a-b\right)}=0\)
\(\Rightarrow-\frac{b-a}{\left(c-a\right)\left(b-c\right)}-\frac{c-a}{\left(a-b\right)\left(b-c\right)}-\frac{c-b}{\left(c-a\right)\left(a-b\right)}=0\)
\(\Rightarrow-\left[\frac{b+a}{\left(c-a\right)\left(b-c\right)}+\frac{c+a}{\left(a-b\right)\left(b-c\right)}+\frac{c+b}{\left(c-a\right)\left(a-b\right)}\right]=0\)
\(\Rightarrow-\left[\frac{\left(b+a\right)\left(a-b\right)^2\left(b-c\right)\left(c-a\right)+\left(c+a\right)\left(c-a\right)^2\left(b-c\right)\left(a-b\right)+\left(c+b\right)\left(b-c\right)^2\left(c-a\right)\left(a-b\right)}{\left(b-c\right)^2\left(c-a\right)^2\left(a-b\right)^2}\right]=0\)
\(\Rightarrow-\left\{\frac{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left[\left(b+a\right)\left(a-b\right)+\left(c+a\right)\left(c-a\right)+\left(b+c\right)\left(b-c\right)\right]}{\left(b-c\right)^2\left(c-a\right)^2\left(a-b\right)^2}\right\}=0\)
\(\Rightarrow-\left[\frac{\left(b+a\right)\left(b-a\right)+\left(c+a\right)\left(c-a\right)+\left(b+c\right)\left(b-c\right)}{\left(b-c\right)\left(c-a\right)\left(a-b\right)}\right]=0\)
\(\Rightarrow-\left[\frac{\left(a^2-b^2\right)+\left(c^2-a^2\right)+\left(b^2-c^2\right)}{\left(b-c\right)\left(c-a\right)\left(a-b\right)}\right]=0\)
\(\Rightarrow-\left[\frac{\left(-b^2+b^2\right)+\left(-a^2+a^2\right)+\left(-c^2+c^2\right)}{\left(b-c\right)\left(c-a\right)\left(a-b\right)}\right]=0\)
\(\Rightarrow-\left[\frac{0}{\left(b-c\right)\left(c-a\right)\left(a-b\right)}\right]=0\)
\(\Rightarrow0=0\) ( đpcm )
30 tháng 8 2019
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 tháng 10 2017
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)
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.