Ko bao gio nha ban

Giúp mình đi. Sao lại nói ko bao giờ

ok bạn 

hôm nay à

Bài này chỉ vận dụng phân tích đa thức thành nhân tử thôi

Có: \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=6xyz\)

\(\Leftrightarrow2\left(x^2+y^2+z^2-xy-yz-xz\right)=6xyz\)

\(\Leftrightarrow x^2+y^2+z^2-xy-yz-xz=3xyz\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)=3xyz\left(x+y+z\right)\)

\(\Leftrightarrow x^3+y^3+z^3-3xyz=3xyz\left(x+y+z\right)\)

\(\Leftrightarrow x^2+y^3+z^3=3xyz\left(x+y+z+1\right)\)

Do đó: \(x^3+y^3+z^3+1=3xyz\left(x+y+z+1\right)+1⋮x+y+z+1\)

Suy ra: \(1⋮x+y+z+1\)

 \(\Rightarrow x+y+z+1=1\)( do \(x,y,z\ge0\Rightarrow x+y+z+1\ge1\))

\(\Leftrightarrow x=y=z=0\)

Vậy \(x=y=z=0\)

Ai đó giúp pls

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:

 A_abc = \(\frac{abc\left(a-b\right)}{c}+\frac{abc\left(b-c\right)}{a}+\)\(\frac{abc\left(c-a\right)}{b}\)

A_abc =  ab.( a - b ) + bc.( b - c ) + ac.( c - a )

A_abc = ab.( a - b ) + bc.( a - c + b - a ) + ac.( a - c )

A_abc = ab.( a - b ) - bc.( a - b ) - bc.( c - a ) + ac.(c - a )

A_abc = b.( a - b ).( a - c ) - c.( a - b ).(c - a ) 

A_abc= ( 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

\(\Rightarrow\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right).\)\(\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)=9\)

MÌNH CŨNG KHÔNG CHẮC LẮM !