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

c: \(=\dfrac{\left(x-5\right)\left(x+5\right)}{3x+4}\cdot\dfrac{-5}{x-5}=\dfrac{-5\left(x+5\right)}{3x+4}\)

\(a,\left(x-1\right)\left(2x-1\right)\)

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

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

\(b,\left(9x^4+12x^3-15x^2-3x\right):3x\)

\(=3x^3+4x^2-5x-1\)

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

\(=x^3+4x^2+5x-x^3-4x^2\)

\(=5x\)

b) \(\left(x-2\right)^2+\left(3-x\right)\left(x-1\right)\)

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

\(=1\)

c) \(\left(x+2\right)^3-x\left(x^2+6x+12\right)\)

\(=x^3+6x^2+12x+8-x^3-6x^2-12x\)

\(=8\)

Ta có: 

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

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

\(=\dfrac{1}{x+1}\)