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

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

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

Ta có: \(\left(x-2\right)^2+\left(3-x\right)\left(x-1\right)\)

\(=\left(x-2\right)^2-\left(x-3\right)\left(x-1\right)\)

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

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

=4-3

=1

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

Bài 3:

3: \(6x\left(x-y\right)-9y^2+9xy\)

\(=6x\left(x-y\right)+9xy-9y^2\)

\(=6x\left(x-y\right)+9y\left(x-y\right)\)

\(=\left(x-y\right)\left(6x+9y\right)\)

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

Bài 4:

$(x-1)(x^2+x+1)-x^3-6x=11$

Dùng $(x-1)(x^2+x+1)=x^3-1$:

$x^3-1-x^3-6x=11$

$-6x-1=11$

$-6x=12$

$x=-2$

Vậy $x=-2$.

$16x^2-(3x-4)^2=0$

$(4x)^2-(3x-4)^2=0$

$(4x-3x+4)(4x+3x-4)=0$

$(x+4)(7x-4)=0$

$x=-4$ hoặc $x=\dfrac47$

Vậy $x=-4,\dfrac47$.

$x^3-x^2+3-3x=0$

$=x^2(x-1)-3(x-1)=0$

$=(x-1)(x^2-3)=0$

$x-1=0$ hoặc $x^2-3=0$

$x=1$ hoặc $x=\pm\sqrt3$

Vậy $x=1,\sqrt3,-\sqrt3$.

$\dfrac{x-1}{x+2}=\dfrac{x+2}{x+1}$

Điều kiện: $x\ne-2,-1$.

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

$x^2-1=x^2+4x+4$

$-4x=5$

$x=-\dfrac54$

Vậy $x=-\dfrac54$.

$\dfrac1{x+2}+\dfrac2{x+1}=0$

Điều kiện: $x\ne-2,-1$.

$\dfrac{x+1+2(x+2)}{(x+2)(x+1)}=0$

$x+1+2x+4=0$

$3x+5=0$

$x=-\dfrac53$

Vậy $x=-\dfrac53$.

$\dfrac{9-x^2}{x}:(x-3)=1$

Điều kiện: $x\ne0,3$.

$\dfrac{9-x^2}{x(x-3)}=1$

$9-x^2=x(x-3)$

$9-x^2=x^2-3x$

$2x^2-3x-9=0$

$(2x+3)(x-3)=0$

$x=-\dfrac32$ hoặc $x=3$

Nhưng $x=3$ không thỏa điều kiện.

Vậy $x=-\dfrac32$.

Nhận thấy \(x^3-x=x\left(x^2-1\right)=x\left(x-1\right)\left(x+1\right)\)

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

Ta có: \(\left(x^3+4x^2+x-2\right):\left(x+1\right)\)

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

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