\(\left(2x+1\right)\left(x+1\right)^2\left(2x+3\right)=18\)
\(\Leftrightarrow\left[\left(2x+1\right)\left(2x+3\right)\right]\left(x^2+2x+1\right)=18\)
\(\Leftrightarrow4\cdot\left(4x^2+8x+3\right)\left(x^2+2x+1\right)=4\cdot18\)
\(\Leftrightarrow\left(4x^2+8x+3\right)\left(4x^2+8x+4\right)=72\)
Đặt \(4x^2+8x+3=a\)
\(\Leftrightarrow a\left(a+1\right)=72\)
\(\Leftrightarrow a^2+a-72=0\)
\(\Leftrightarrow a^2+9a-8a-72=0\)
\(\Leftrightarrow a\left(a+9\right)-8\left(a+9\right)=0\)
\(\Leftrightarrow\left(a+9\right)\left(a-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-9\\a=8\end{matrix}\right.\)
TH1 : \(a=-9\)
\(\Leftrightarrow4x^2+8x+3=-9\)
\(\Leftrightarrow4x^2+8x+12=0\)
\(\Leftrightarrow4\left(x^2+2x+3\right)=0\)
\(\Leftrightarrow x^2+2x+1+2=0\)
\(\Leftrightarrow\left(x+1\right)^2=-2\)
=> vô nghiệm
TH2 : \(a=8\)
\(\Leftrightarrow4x^2+8x+3=8\)
\(\Leftrightarrow4x^2+8x-5=0\)
\(\Leftrightarrow\left(2x\right)^2+2\cdot2x\cdot2+2^2-9=0\)
\(\Leftrightarrow\left(2x+2\right)^2=9=\left(\pm3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+2=3\\2x+2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-5}{2}\end{matrix}\right.\)
Vậy....
Đề đúng: (2x + 1)(x + 1)2(4x + 6) = 18
\(\left(2x+1\right)\left(2x+3\right)\left(x+1\right)^2=18\)
\(\Leftrightarrow\left(2x+2-1\right)\left(2x+2+1\right)\left(x+1\right)^2=18\)
\(\Leftrightarrow\left(\left(2x+2\right)^2-1\right)\left(x+1\right)^2=18\)
\(\Leftrightarrow4\left(x+1\right)^4-\left(x+1\right)^2-18=0\)
Đặt t = \(\left(x+1\right)^2\) \(\left(t\ge0\right)\)
Phương trình \(\Leftrightarrow4t^2-t-18=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{9}{4}\left(nh\right)\\t=-2\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\left(x+1\right)^2-\dfrac{9}{4}=0\)
\(\Leftrightarrow\left(x+1-\dfrac{3}{2}\right)\left(x+1+\dfrac{3}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
