(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

\(\frac{2x}{x+1}+\frac{18}{x^2+2x-3}=\frac{2x-5}{x+3}ĐKXĐ:x\ne-1;-3\)

\(\frac{2x}{x+1}+\frac{18}{\left(x-1\right)\left(x+3\right)}=\frac{2x-5}{x+3}\)

\(2x\left(x-1\right)\left(x+3\right)+18\left(x+1\right)=\left(2x-5\right)\left(x+1\right)\left(x-1\right)\)

\(4x^2+12x+18=-2x-5x^2+5\)

\(4x^2+12x+18+2x+5x^2-5=0\)

\(9x^2-14x+13=0\)

=> vô nghiệm

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge1\\-4+\sqrt{7}\le x\le-1\end{matrix}\right.\)

Khi x thỏa ĐKXĐ, vế phải luôn dương, bình phương 2 vế ta được:

\(\Leftrightarrow3x^2+16x+17+2\sqrt{\left(x^2-1\right)\left(2x^2+16x+18\right)}=4x^2+16x+16\)

\(\Leftrightarrow2\sqrt{\left(x^2-1\right)\left(2x^2+16x+18\right)}=x^2-1\)

\(\Leftrightarrow4\left(x^2-1\right)\left(2x^2+16x+18\right)=\left(x^2-1\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=0\\4\left(2x^2+16x+18\right)=x^2-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm1\\7x^2+64x+73=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm1\\x=\dfrac{-32+3\sqrt{57}}{7}\\x=\dfrac{-32-3\sqrt{57}}{7}\left(loại\right)\end{matrix}\right.\)

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

\(\Leftrightarrow\left(2x+1\right)\left(2x+3\right)\left(x^2+2x+1\right)-18=0\)

\(\Leftrightarrow\left(4x^2+8x+3\right)\left(x^2+2x+1\right)-18=0\)

\(\Leftrightarrow4\left(x^2+2x+\frac{3}{4}\right)\left(x^2+2x+1\right)-18=0\)

Đặt \(a=x^2+2x+\frac{3}{4}\)    \(a=x^2+2x+\frac{3}{4}\)

\(\Rightarrow4a\left(a+\frac{1}{4}\right)-18=0\)

\(\Leftrightarrow4a^2+a-18=0\)

\(\Leftrightarrow4a^2-8a+9a-18=0\)

\(\Leftrightarrow\left(4a+9\right)\left(a-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4a+9=0\\a-2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}a=-\frac{9}{4}\\a=2\end{cases}}\)

\(\left(+\right)a=-\frac{9}{4}\Rightarrow x^2+2x+\frac{3}{4}=-\frac{9}{4}\)

\(\Leftrightarrow x^2+2x+\frac{3}{4}+\frac{9}{4}=0\)\(\Leftrightarrow x^2+2x+3=0\)

\(\Leftrightarrow\left(x+1\right)^2+2=0\)

( vô lí )

\(\left(+\right)a=2\Rightarrow x^2+2x+\frac{3}{4}=2\)

\(\Leftrightarrow x^2+2x-\frac{5}{4}=0\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{5}{2}=0\\x-\frac{1}{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{2}\\x=\frac{1}{2}\end{cases}}}\)

=> (2x+1)(2x+3)(x+1)²=18

=> (2x+2-1)(2x+2+1)(x+1)²=18

=> ((2x+2)²-1)(x+1)²=18

=>(2x+2)²(x+1)^{2 _} (x+1)² - 18 =0

=> (2(x+1))²(x+1)^2_(x+1)² - 18=0

=> 4(x+1)⁴ - (x+1)² -18 =0

 đặt (x+1)²=a

phương trình <=> 4a² - a-18=0

=>  4a² + 8a - 9a -18=0

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

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

từ đó tìm ra a xong tìm ra x mình nghĩ bạn giải đc :D 

a: =>4x-3x=1-2

=>x=-1

b: =>3x=12

=>x=4

c: =>2(x^2-6)=x(x+3)

=>2x^2-12-x^2-3x=0

=>x^2-3x-12=0

=>\(x=\dfrac{3\pm\sqrt{57}}{2}\)