Giải:

Thay \(x-y=9\) vào biểu thức ta được:

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

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

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

Vậy \(A=0\)

Từ x - y = 9 \(\Rightarrow\) x=9+y.

Thay x=9+y vào A ta có:

A=\(\dfrac{4x-9}{3x+y}\)-\(\dfrac{4y+9}{3y+x}\)

=\(\dfrac{4.\left(9+y\right)-9}{3\left(9+y\right)+y}\)-\(\dfrac{4y+9}{3y+y+9}\)

=\(\dfrac{36+4y-9}{27+3y+y}\)-\(\dfrac{4y+9}{4y+9}\)

=\(\dfrac{27+4y}{27+4y}\)-1=1-1=0

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

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

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

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

Thay 9 = x - y vào biểu thức B , ta được :

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

Vậy ...

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

Thay x-y=9 vào biểu thức (1),được:

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

B=3x+y/3x+y-3y+x/3y+x

B=1-1

B=0

Theo đề bài, ta có: x-y=9 => x=9+y

Thay x=9+y vào biểu thức B, ta có:

\(\dfrac{4.\left(9+y\right)-9}{3.\left(9+y\right)+y}\)-\(\dfrac{4y+9}{3y+\left(9+y\right)}\)

=\(\dfrac{36+4y-9}{27+3y+y}\)-\(\dfrac{4y+9}{4y+9}\)

=\(\dfrac{27+4y}{27+4y}\)-\(1\)

=\(1-1\)=\(0\)

Vậy....

x+y=9 nên x=9-y

\(M=\dfrac{4\left(9-y\right)-9}{3\left(9-y\right)+y}-\dfrac{4y+9}{3y+9-y}\)

\(=\dfrac{36-4y-9}{27-3y+y}-\dfrac{4y+9}{2y+9}\)

\(=\dfrac{4y-27}{2y-27}-\dfrac{4y+9}{2y+9}\)

\(=\dfrac{8y^2+36y-54y-243-\left(8y^2-108y+18y-243\right)}{\left(2y-27\right)\left(2y+9\right)}\)

\(=\dfrac{8y^2-18y-243-8y^2+90y+243}{\left(2y-27\right)\left(2y+9\right)}=\dfrac{72y}{\left(2y-27\right)\left(2y+9\right)}\)

x-y=9=>x=y+9,thay x=y+9 vào B ta có:

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

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

Vậy B=0

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

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

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

\(x-y=9\Rightarrow x=9+y\)

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

 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