B=(4x-9)/(3x+y)-(4y+9)/(3y+x) 
= [4x-(x-y)]/(3x+y) - [4y+(x-y)]/(3y+x) 
= (4x-x+y)/(3x+y) - (4y+x-y)/(3y+x) 
= (3x+y)/(3x+y) - (3y+x)/(3y+x) 
= 1 - 1 = 0

x - y = 9

=> x = 9 + y

=> B = \(\frac{4.\left(9+y\right)-9}{3.\left(9+y\right)+y}-\frac{4.y+9}{3.y+\left(9+y\right)}=\frac{36+4y-9}{27+3y+y}-\frac{4y+9}{4y+9}=\frac{27+4y}{27+4y}-1=1-1=0\)

Mình tưởng là \(B=\frac{4x-9}{3x+y}-\frac{4y+9}{3y+x}\)chứ nhỉ (vòng 11 violympic phải không bạn?)

Thay \(x-y=9\)vào biểu thức A ta được:

\(A=\frac{4x-\left(x-y\right)}{3x+y}-\frac{4y+\left(x-y\right)}{3y+x}=\frac{4x-x+y}{3x+y}-\frac{4y+x-y}{3y+x}\)

   \(=\frac{3x+y}{3x+y}-\frac{3y+x}{3y+x}=1-1=0\)

B=(4x - 9 )/(3x + y )-(4y +9)/(3y+x)

  =[4x - (x-y)]/(3x + y ) - ([4y + (x-y)]/(3y + x)

 =[4x-x + y)/(3x + y ) - [4y +(x-y)]/(3y + x) 

 =(3x + y )/(3x + y)- ( 3y + x )/(3y+x)

 = 1 -1 

=0

VAY :   B=0

thay x-y =9 vào biểu thức B = \(\frac{4x-9}{3x+y}-\frac{4y+9}{3y+x}\) , ta có 

B = \(\frac{4x-\left(x-y\right)}{3x+y}-\frac{4y+\left(x-y\right)}{3y+x}\)

   = \(\frac{4x-x+y}{3x+y}-\frac{4y+x-y}{3y+x}\)

    = \(\frac{3x+y}{3x+y}-\frac{3y+x}{3y+x}\)

    = 1 - 1

    = 0

vậy với x-y=9 thì giá trị biểu thức  B = \(\frac{4x-9}{3x+y}-\frac{4y+9}{3y+x}\) là 0

Ta có : \(x-y=9\)  => \(y=x-9\)         ;  \(x=y+9\)

=> \(B=\frac{4x-9}{3x+y}-\frac{4y+9}{3y+x}=\frac{4x-9}{3x+x-9}-\frac{4y+9}{3y+y+9}\)

=> \(B=\frac{4x-9}{4x-9}-\frac{4y+9}{4y+9}=1-1=0\)

=> \(B=0\)

 B=(4x-9)/(3x+y)-(4y+9)/(3y+x) 
= [4x-(x-y)]/(3x+y) - [4y+(x-y)]/(3y+x) 
= (4x-x+y)/(3x+y) - (4y+x-y)/(3y+x) 
= (3x+y)/(3x+y) - (3y+x)/(3y+x) 
= 1 - 1 = 0

x - y = 9 => x = 9 + y thay vào B ta được :

\(B=\frac{4\left(9+y\right)-9}{3\left(9+y\right)+y}-\frac{4y+9}{3y+9+y}=\frac{36+4y-9}{27+3y+y}-\frac{4y+9}{4y+9}=\frac{27+4y}{27+4y}-\frac{4y+9}{4y+9}=1-1=0\)

Vậy B = 0