(2x+1)(x+1)2(2x+3)-18=0
\(\Leftrightarrow\)(2x+1)(x+1)2(2x+3)=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 \(t=\left(x+1\right)^2\left(t\ge0\right)\)
\(\Leftrightarrow4t^2-t-18=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{9}{4}\left(tm\right)\\t=-2\left(ktm\right)\end{matrix}\right.\)
\(\Leftrightarrow\left(x+1\right)^2-\dfrac{9}{4}=0\)
\(\Leftrightarrow\left(x+1-\dfrac{2}{3}\right)\left(x+1+\dfrac{2}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
\(\left(2x+1\right)\left(x+1\right)^2\left(2x+3\right)-18=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x+3\right)\left(x^2+2x+1\right)-18=0\)
\(\Leftrightarrow\left(4x^2+6x+2x+3\right)\left(x^2+2x+1\right)-18=0\)
\(\Leftrightarrow\left(4x^2+8x+3\right)\left(x^2+2x+1\right)-18=0\)
\(\Leftrightarrow\left(4x^2+8x+3\right).4.\left(x^2+2x+1\right)-4.18=0\)
\(\Leftrightarrow\left(4x^2+8x+3\right)\left(4x^2+8x+4\right)-72=0\)
-Đặt \(t=4x^2+8x+3\)
PT\(\Leftrightarrow t\left(t+1\right)-72=0\)
\(\Leftrightarrow t^2+t-72=0\)
\(\Leftrightarrow t^2-8t+9t-72=0\)
\(\Leftrightarrow t\left(t-8\right)+9\left(t-8\right)=0\)
\(\Leftrightarrow\left(t-8\right)\left(t+9\right)=0\)
\(\Leftrightarrow t-8=0\) hay \(t+9=0\)
\(\Leftrightarrow4x^2+8x+3-8=0\) hay \(4x^2+8x+3+9=0\)
\(\Leftrightarrow4x^2+8x-5=0\) hay \(4x^2+8x+12=0\)
\(\Leftrightarrow4x^2-2x+10x-5=0\) hay \(\left(2x\right)^2+2.2x.2+4+8=0\)
\(\Leftrightarrow2x\left(2x-1\right)+5\left(2x-1\right)=0\) hay \(\left(2x+2\right)^2+8=0\) (phương trình vô nghiệm vì \(\left(2x+2\right)^2+8\ge8\))
\(\Leftrightarrow\left(2x-1\right)\left(2x+5\right)=0\)
\(\Leftrightarrow2x-1=0\) hay \(2x+5=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\) hay \(x=\dfrac{-5}{2}\)
-Vậy \(S=\left\{\dfrac{1}{2};\dfrac{-5}{2}\right\}\)
