Sửa đề: \(\frac{x^2+3xy}{x^2-9y^2}+\frac{2x^2-5xy-3y^2}{x^2-6xy+9y^2}=\frac{3x^2+xy+3xz+yz}{x^2-3yz+xz-3xy}\)

\(\frac{x^2+3xy}{x^2-9y^2}+\frac{2x^2-5xy-3y^2}{x^2-6xy+9y^2}\)

\(=\frac{x\left(x+3y\right)}{\left(x-3y\right)\left(x+3y\right)}+\frac{2x^2-6xy+xy-3y^2}{\left(x-3y\right)^2}\)

\(=\frac{x}{x-3y}+\frac{\left(x-3y\right)\left(2x+y\right)}{\left(x-3y\right)^2}=\frac{x}{x-3y}+\frac{2x+y}{x-3y}=\frac{3x+y}{x-3y}\)

\(\frac{3x^2+xy+3xz+yz}{x^2-3yz+xz-3xy}\)

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

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

Do đó: \(\frac{x^2+3xy}{x^2-9y^2}+\frac{2x^2-5xy-3y^2}{x^2-6xy+9y^2}=\frac{3x^2+xy+3xz+yz}{x^2-3yz+xz-3xy}\)

\(\frac{4.\left(x+3\right)}{3x-1}:\frac{x^2+3x}{3x-1}=\frac{4.\left(x+3\right)}{\left(3x-1\right)}\cdot\frac{\left(3x-1\right)}{x^2+3x}=\frac{4.\left(x+3\right)}{x.\left(x+3\right)}=\frac{4}{x}\)

\(a,\frac{x+2}{x-1}-\frac{x-9}{1-x}-\frac{x-9}{1-x}\)

\(=\frac{-x-2}{1-x}-\frac{x-9}{1-x}-\frac{x-9}{1-x}\)

\(=\frac{-x-2}{1-x}+\frac{-\left(x-9\right)}{1-x}+\frac{-\left(x-9\right)}{1-x}\)

\(=\frac{-x-2-x+9-x+9}{1-x}=\frac{-3x+16}{1-x}\)

Câu b,c mk chưa học, bn thông cảm

Còn câu a, nếu sai thì xin lượng thứ :))

Lời giải:
a)

\(\frac{x+2}{x-1}-\frac{x-9}{1-x}-\frac{x-9}{1-x}=\frac{x+2}{x-1}-\frac{2(x-9)}{1-x}\)

\(=\frac{x+2}{x-1}+\frac{2(x-9)}{x-1}=\frac{x+2+2(x-9)}{x-1}=\frac{3x-16}{x-1}\)

b)

\(\frac{x^2-9y^2}{x^2y}: \frac{xz-3yz}{3xy}=\frac{x^2-9y^2}{x^2y}.\frac{3xy}{xz-3yz}\)

\(=\frac{(x-3y)(x+3y)}{x^2y}.\frac{3xy}{z(x-3y)}=\frac{3(x+3y)}{xz}\)

c) \(\frac{4(x+3)}{3x-1}:\frac{x^2+3x}{3x-1}=\frac{4(x+3)}{3x-1}.\frac{3x-1}{x^2+3x}=\frac{4(x+3)}{x^2+3x}=\frac{4(x+3)}{x(x+3)}=\frac{4}{x}\)

\(\dfrac{x+9y}{x^2-9y^2}-\dfrac{3y}{x^2+3xy}\)

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

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

\(=\dfrac{x^2-6xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}\)

\(=\dfrac{\left(x-3y\right)^2}{x\left(x-3y\right)\left(x+3y\right)}\)

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

b: \(=\dfrac{-1}{x\left(5x-1\right)}-\dfrac{25x-15}{\left(5x-1\right)\left(5x+1\right)}\)

\(=\dfrac{-5x-1-25x^2+15x}{x\left(5x-1\right)\left(5x+1\right)}\)

\(=\dfrac{-25x^2-10x-1}{x\left(5x-1\right)\left(5x+1\right)}=\dfrac{-\left(5x+1\right)}{x\left(5x-1\right)}\)

c: \(=\dfrac{x+9y}{\left(x-3y\right)\left(x+3y\right)}-\dfrac{3y}{x\left(x-3y\right)}\)

\(=\dfrac{x^2+9xy-3xy-9y^2}{x\left(x-3y\right)\left(x+3y\right)}\)

\(=\dfrac{x^2+6xy-9y^2}{x\left(x-3y\right)\left(x+3y\right)}\)

d: \(=\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{3x^2+4x+1-x^2+2x-1+x^2+2x-3}{\left(x-1\right)^2\cdot\left(x+1\right)}\)

\(=\dfrac{3x^2+8x-3}{\left(x-1\right)^2\cdot\left(x+1\right)}=\dfrac{3x^2+9x-x-3}{\left(x-1\right)^2\cdot\left(x+1\right)}\)

\(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}\)

viết đầu bài rõ ràng 1 chút chả hiểu gì cả

Ta có: M=\((\frac{1}{3}x-y)(x^2+3xy+9y^2)+(9y^3-\frac{1}{3}x^3) \)

=\((\frac{1}{3}x-y)(x^2+3xy+9y^2)-\frac{1}{3}(3y-x)(9y^2+3xy+x^2) \)

=\((\frac{1}{3}x-y)(x^2+3xy+9y^2)-(\frac{1}{3}x-y)(x^2+3xy+9y^2)=0\)

=> M không phụ thuộc vào x,y