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

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

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

\(=8\)

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

2: 

a: =>-2x=10

=>x=-5

b: =>(x-3)(2x+5)=0

=>x=3 hoặc x=-5/2

\(=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{13\cdot15}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{13}-\dfrac{1}{15}=\dfrac{14}{15}\)

a/ \(\left(2x+3\right)\left(x-5\right)-\left(x-1\right)^2+x\left(7-x\right)\)

\(=2x^2-2x-15-x^2+2x-1+7x-x^2\)

\(=7x-16\)

b, = x² - 16 - ( x³ - 3³ ) : ( x - 3 )

= x² - 16 - \([\) ( x - 3 ) ( x² + 3x + 9 ) \(]\) : ( x - 3 )

= x² - 16 - ( x² + 3x + 9 )

= x² - 16 - x² - 3x - 9

= -25 - 3x

\(a,=2x^2-10x+3x-15-x^2+2x-1+7x-x^2=2x-16\\ b,=x^2-16-\left(x-3\right)\left(x^2+3x+9\right):\left(x-3\right)\\ =x^2-16-x^2-3x-9=-3x-25\)

\(a.4\)

\(b.9\)

Ta nhóm các chữ số với nhau: (tận cùng hai với tận cùng năm) và các thừa số tận cùng 0