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

\(\Leftrightarrow\left(2x-1\right)\left(2x-3\right)\left(2x+2\right)^2=72\) (*)

Đặt \(a=2x+2\)

(*) \(\Leftrightarrow\left(a-3\right)\left(a-5\right).a^2=0\)

\(\Leftrightarrow\left(a^3-5a^2\right)\left(a-3\right)=0\)

\(\Leftrightarrow a^4-8a^3+15a^2=0\)

\(\Leftrightarrow a^4-5a^3-3a^3+15a^2=0\)

\(\Leftrightarrow a^3.\left(a-5\right)-3a^2.\left(a-5\right)=0\)

\(\Leftrightarrow\left(a-5\right)\left(a-3\right).a^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x-1=0\\\left(2x+2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)

Đặt x² + 2x = a ta có

\(\frac{1}{a-3}\)+ \(\frac{18}{a+2}\)= \(\frac{18}{a+1}\)

<=> a² - 15a + 56 = 0

<=> a = (7;8)

Thế vô tìm được nghiệm 

x=-0,384367156686985

x=0,442125301696298

x=2,9422181027264

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

\(\Leftrightarrow\)\(\left(2x+1\right)\left(x+1\right)^2.4.\left(2x+3\right)=72\)

\(\Leftrightarrow\)\(\left(2x+1\right)\left(2x+2\right)^2\left(2x+3\right)=72\)

Đặt               \(2x+2=a\)     ta có:

      \(\left(a-1\right)a^2\left(a+1\right)=72\)

\(\Leftrightarrow\)\(\left(a^2-1\right)a^2-72=0\)

\(\Leftrightarrow\)\(a^4-a^2-72=0\)

\(\Leftrightarrow\)\(\left(a^2+8\right)\left(a^2-9\right)=0\)

\(\Leftrightarrow\)\(\left(a^2+8\right)\left(a-3\right)\left(a+3\right)=0\)

Vì               \(a^2+8>0\)   \(\forall a\)

\(\Rightarrow\)\(\orbr{\begin{cases}a-3=0\\a+3=0\end{cases}}\)

Thay trở lại ta được:        \(\orbr{\begin{cases}2x-1=0\\2x+5=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0,5\\x=-2,5\end{cases}}\)

Vậy...

P/S:   tham khảo nhé!!!!   chúc bạn học tốt   ^_^

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

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

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

đặt (x^2+2x)=Pain ta được

\(\left(4pain+3\right)\left(pain+1\right)-18=0\)

\(4pain^2+4pain+3pain+3-18=0\)

\(4pain^2+7pain-15=0\)

\(4pain^2+12pain-5pain-15=0\)

\(4pain\left(pain+3\right)-5\left(pain+3\right)=0\)

\(\left(Pain+3\right)\left(4pain-5\right)=0\)

\(Pain=-3;Pain=\frac{5}{4}\)

rồi m đến đây tự làm đi nhé 

khó thế

khó thế

\(3\left(x-2\right)-4x+5=2\left(2x+1\right)-18\\ \Leftrightarrow3x-6-4x+5=4x+2-18\\ \Leftrightarrow-x-1=4x-16\\ \Leftrightarrow-x-4x=-16+1\\ \Leftrightarrow-5x=-15\\ \Leftrightarrow x=3\)

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

\(ĐK:x\in R\)

\(pt\left(1\right)\Leftrightarrow2x^2+16x+18+x^2+1+2\sqrt[]{(2x^2+16x+18)\left(x^2+1\right)}=4x^2+16x+16\)

\(\Leftrightarrow3+2\sqrt{(2x^2+16x+18)\left(x^2+1\right)}=x^2\)

\(\Leftrightarrow2\sqrt{(2x^2+16x+8)\left(x^2+1\right)}=x^2-3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-3\ge0\\4\left(2x^2+16x+8\right)\left(x^2+1\right)=x^4-6x^2+9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-\sqrt{3}\le x\le\sqrt{3}\\4\left(2x^4+16x^3+10x^2+16x+8\right)=x^4-6x^2+9\end{matrix}\right.\)

\(\Leftrightarrow7x^4+64x^3+46x^2+64x+23=0\)