9+y*y-3y
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x+y=9 nên x=9-y
\(M=\dfrac{4\left(9-y\right)-9}{3\left(9-y\right)+y}-\dfrac{4y+9}{3y+9-y}\)
\(=\dfrac{36-4y-9}{27-3y+y}-\dfrac{4y+9}{2y+9}\)
\(=\dfrac{4y-27}{2y-27}-\dfrac{4y+9}{2y+9}\)
\(=\dfrac{8y^2+36y-54y-243-\left(8y^2-108y+18y-243\right)}{\left(2y-27\right)\left(2y+9\right)}\)
\(=\dfrac{8y^2-18y-243-8y^2+90y+243}{\left(2y-27\right)\left(2y+9\right)}=\dfrac{72y}{\left(2y-27\right)\left(2y+9\right)}\)
\(\dfrac{y}{y+3}-\dfrac{2y}{3-y}-\dfrac{3y^2+9}{y^2-9}=\dfrac{y\left(y-3\right)}{\left(y-3\right)\left(y+3\right)}+\dfrac{2y\left(y+3\right)}{\left(y-3\right)\left(y+3\right)}-\dfrac{3y^2+9}{\left(y-3\right)\left(y+3\right)}=\dfrac{y^2-3y+2y^2+6y-3y^2-9}{\left(y-3\right)\left(y+3\right)}=\dfrac{3y-9}{\left(y-3\right)\left(y+3\right)}=\dfrac{3\left(y-3\right)}{\left(y-3\right)\left(y+3\right)}=\dfrac{3}{y+3}\)
\(=\dfrac{y^2-3y+2y^2+6y-3y^2+9}{\left(y-3\right)\left(y+3\right)}=\dfrac{3y+9}{\left(y-3\right)\left(y+3\right)}=\dfrac{3}{y-3}\)
Cho x-y=9 => x=9+y thay vào B ta được:
B=4(9+y)-9/3(9+y)+y - 4y+9/3y+9+y
B= 36+4y-9/27+4y - 4y+9/4y+9
B= 37+4y/27+4y - 4y+9/4y+9
B= 1-1
B=0
Thay \(x-y=9\)vào biểu thức A ta được:
\(A=\frac{4x-\left(x-y\right)}{3x+y}-\frac{4y+\left(x-y\right)}{3y+x}=\frac{4x-x+y}{3x+y}-\frac{4y+x-y}{3y+x}\)
\(=\frac{3x+y}{3x+y}-\frac{3y+x}{3y+x}=1-1=0\)