\(2x\left(x-3\right)-2x^2=4\\ \Leftrightarrow2x^2-6x-2x^2=4\\ \Leftrightarrow-6x=4\\ \Leftrightarrow x=-\dfrac{2}{3}\\ KL:...\)

\(\left(2x+1\right)\left(2x+3\right)\left(x+1\right)^2=18\)

\(\Leftrightarrow\left(2x+2-1\right)\left(2x+2+1\right)\left(x+1\right)^2=18\)

\(\Leftrightarrow\left(\left(2x+2\right)^2-1\right)\left(x+1\right)^2=18\)

\(\Leftrightarrow4\left(x+1\right)^4-\left(x+1\right)^2-18=0\)

Đặt t = \(\left(x+1\right)^2\) \(\left(t\ge0\right)\)

pt \(\Leftrightarrow4t^2-t-18=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{9}{4}\left(nh\right)\\t=-2\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow\left(x+1\right)^2-\dfrac{9}{4}=0\)

\(\Leftrightarrow\left(x+1-\dfrac{3}{2}\right)\left(x+1+\dfrac{3}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

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

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

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

=> x²+2x-x+2=x²+3

<=>x=3

(2x+1)(x+1)²(2x+3)=18 

<=> (2x+2-1)(x+1)²(2x+2+1)=18

Đặt y=x+1, ta có: 

(2y-1)y²(2y+1)=18

Ta có 

(2x+1)(x+1)²(2x+3)=18

=> (x+1)²(4x²+8x+3)-18=0

=> (x²+2x+1)(4x²+8x+3)-18=0

Đặt x²+2x+1=a ta có 

a.(4a-1)-18=0

=> 4a²-a-18=0

=> 4a² +8a-9a-18=0

=> 4a(a+2)-9(a+2)=0

=> (a+2)(4a-9)=0

Với a=x²+2x+1biểu thức trên trở thành

(x²+2x+3)(4x²+8x-5)=0

=> x²+2x+3=0 hoặc 4x²+8x-5=0

• x²+2x+3=0 => phương trình vô nghiệm

• 4x²+8x-5=0 => x=1/2 hoặc x=-5/2

Vậy x=1/2 và x=-5/2 là nghiệm của phương trình

ĐKXĐ:\(x\ne\pm1\)

\(\dfrac{4x+5}{x-1}+\dfrac{2x-1}{x+1}=6\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(4x+5\right)}{\left(x+1\right)\left(x-1\right)}+\dfrac{\left(x-1\right)\left(2x-1\right)}{\left(x+1\right)\left(x-1\right)}=6\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(4x+5\right)+\left(x-1\right)\left(2x-1\right)}{\left(x+1\right)\left(x-1\right)}=6\)

\(\Leftrightarrow4x^2+4x+5x+5+2x^2-2x-x+1=6\left(x^2-1\right)\\ \Leftrightarrow6x^2+6x+6=6x^2-6\\ \Leftrightarrow6x=-12\\ \Leftrightarrow x=-2\left(tm\right)\)

\(\dfrac{4x+5}{x-1}+\dfrac{2x-1}{x+1}=6\)

\(\dfrac{\left(4x+5\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{\left(2x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(4x+5\right)\left(x+1\right)+\left(2x-1\right)\left(x-1\right)}{x^2-1}\)

\(\dfrac{4x^2+9x+5+2x^2-3x+1}{x^2-1}=\dfrac{6x^2+6x+6}{x^2-1}=6\)

\(\Rightarrow6x^2+6x+6=6\left(x^2-1\right)=6x^2-6\)

\(\Rightarrow6x+12=0\Rightarrow x=-2\)