\(\left(x-3y\right)\left(x^2+3xy+9y^2\right)+\left(x+3y\right)\left(x^2-3xy+9y^2\right)\)

\(=x^3-27y^3+x^3+27y^3=2x^3=2.\left(-1\right)^3=-2\)

\(\Leftrightarrow x^3-3y^3+x^3+3y^3\)

\(\Leftrightarrow2x^3\)

\(A=(x+3y)(x^2-3xy+9y^2)+3y(x+3y)(x-3y)-x(3xy+x^2-5)-5x+1\\A=(x+3y)[x^2-x\cdot3y+(3y)^2]+3y[x^2-(3y)^2]-3x^2y-x^3+5x-5x+1\\A=x^3+(3y)^3+3y(x^2-9y^2)-3x^2y-x^3+1\\A=x^3+27y^3+3x^2y-27y^3-3x^2y-x^3+1\\A=1\)$\Rightarrow$ Giá trị của $A$ không phụ thuộc vào giá trị của biến.

đk:   \(x\ne0\);  \(x\ne\pm3y\)

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

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

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

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

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

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

a: \(\left(2x+3\right)^3=8x^3+36x^2+54x+27\)

b: \(\left(x-3y\right)^3=x^3-9x^2y+27xy^2-27y^3\)

a) \(\left(x^3-2y\right)^3\)

\(=\left(x^3-2y\right)\left[\left(x^3\right)^2+x^3.2y+\left(2y^2\right)\right]\)

\(=\left(x^3-2y\right)\left(x^6+2x^3y+4y^2\right)\)

b) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)

\(=\left(x-3y\right)\left(x^2+x.3y+\left(3y\right)^2\right)\)

\(=\left(x-3y\right)^3\)

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

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

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

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

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

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