Đặt A = \(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\)
B = \(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\)
\(\Rightarrow\)A . B = 9
Ta có : A = \(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\)
Nhân abc với A ta được:
Aabc = \(\frac{abc\left(a-b\right)}{c}+\frac{abc\left(b-c\right)}{a}+\)\(\frac{abc\left(c-a\right)}{b}\)
Aabc = ab.( a - b ) + bc.( b - c ) + ac.( c - a )
Aabc = ab.( a - b ) + bc.( a - c + b - a ) + ac.( a - c )
Aabc = ab.( a - b ) - bc.( a - b ) - bc.( c - a ) + ac.(c - a )
Aabc = b.( a - b ).( a - c ) - c.( a - b ).(c - a )
Aabc= ( a - b ).( a - c ).( b - c )
A = \(\frac{\left(a-b\right).\left(a-c\right).\left(b-c\right)}{abc}\)
Xét a + b + c = 0 \(\Rightarrow\) a + b = - c ; c + a = -b ; b + c = -a
Nhân ( a - b ).( c - a ).( b - c ) với B ta được :
B( a - b).( c - a ).( b - c ) = \(\frac{c\left(a-b\right).\left(c-a\right).\left(b-c\right)}{a-b}\)+ \(\frac{a\left(a-b\right).\left(b-c\right).\left(c-a\right)}{b-c}\)+ \(\frac{b\left(a-b\right).\left(b-c\right).\left(c-a\right)}{c-a}\)
B( a - b ).( c - a ).( b - c ) = c.( c - a ).( b - c ) + a.( b - c ).( c - a ) + b.( a - b ).( b - c)
B( a - b ).( c - a ) .( b - c ) = c.( c - a ).( b - c ) + ( a - b ).( -b - c ).( c - a ) + b.( a - b ).( b - c )
B( a - b ).( c - a ).( b - c ) = c.( c - a ).( b - c ) - b.( a - b ).( c- a ) + b.( a - b ).(b - c ) - c.( a - b ).( c - a )
B( a - b ).( c - a ).( b - c ) = c.( c - a ).( -a + 2b - c ) + b.( a - 2c +b).(a - b )
B( a - b).( c - a ).( b - c ) = -3bc.( b + c - 2a )
B( a - b ).( c - a ).( b - c ) = -9abc
B = \(\frac{9abc}{\left(a-b\right).\left(c-a\right).\left(b-c\right)}\)
NHÂN A VỚI B :
\(\frac{\left(a-b\right).\left(b-c\right).\left(a-c\right)}{abc}\)\(.\)\(\frac{9abc}{\left(a-b\right).\left(b-c\right).\left(c-a\right)}\)= 9
\(\Rightarrow\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right).\)\(\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)=9\)
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