\(=\dfrac{3}{2\left(x+3\right)}+\dfrac{6-x}{2x\left(x+3\right)}=\dfrac{3x+6-x}{2x\left(x+3\right)}=\dfrac{2x+6}{2x\left(x+6\right)}=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}=\dfrac{1}{x}\)

\(=\dfrac{3x+6-x}{2x\left(x+3\right)}=\dfrac{2x+6}{2x\left(x+3\right)}=\dfrac{1}{x}\)

\(=x^2-4x+4+3x-x^2+3-x=-2x+7\)

\(=x^2-4x+4-x^2+2x-3=-2x+1\)

\(\left(x+2\right)^2-\left(x-3\right)\left(x+1\right)=x^2+4x+4-\left(x^2+x-3x-3\right)=x^2+4x+4-x^2-x+3x+3=6x+7.\)

\(\left(5x^4-x^3+3x^2\right)\)

5x⁴-x³+3x²

\(\left(2+x\right)\left(x^2-4\right)-\left(x-2\right)\left(x^2+2x+4\right)=\left(x+2\right)^2\left(x-2\right)-\left(x-2\right)\left(x^2+2x+4\right)=\left(x-2\right)\left[\left(x+2\right)^2-x^2-2x-4\right]=\left(x-2\right)\left(x^2+4x+4-x^2-2x-4\right)=\left(x-2\right)2x=2x^2-4x\)

\(\left(x+2\right)\left(x^2-4\right)-\left(x-2\right)\left(x^2+2x+4\right)\)

\(=x^3-4x+2x^2-8-x^3+8\)

\(=2x^2-4x\)

Bài 1: 

b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)

Bài 2: 

a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)

\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)

d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)

\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)

e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)