1/ \(\Leftrightarrow\left(4x+1\right)\left(3x+2\right)\left(12x-1\right)\left(x+1\right)=28\)
\(\Leftrightarrow\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)=28\)
Đặt \(12x^2+11x+2=t\)
\(\Rightarrow12x^2+11x-1=t-3\)
\(\Rightarrow t\left(t-3\right)=28\Leftrightarrow t^2-3t-28=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=7\\t=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}12x^2+11x+2=7\\12x^2+11x+2=-4\end{matrix}\right.\)
Bạn tự giải nốt và kl
b/ \(\Leftrightarrow\left(x^2+2x+1\right)\left(4x^2+8x+3\right)=18\)
\(\Leftrightarrow\left(4x^2+8x+4\right)\left(4x^2+8x+3\right)=72\)
Đặt \(4x^2+8x+3=t\Rightarrow t+1=4x^2+8x+4\)
\(\Rightarrow t\left(t+1\right)=72\Leftrightarrow t^2+t-72=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=8\\t=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}4x^2+8x+3=8\\4x^2+8x+3=-9\end{matrix}\right.\)
Bạn tự giải và kết luận
Đặt \(2x+2=a\Rightarrow\left\{{}\begin{matrix}2x+1=a-1\\2x+3=a+1\\x+1=\frac{a}{2}\end{matrix}\right.\)
khi đó ta có phương trình:
\(\left(\frac{a}{2}\right)^2\left(a-1\right)\left(a+1\right)=18\Leftrightarrow\frac{a^2}{4}\left(a^2-1\right)=18\)
\(\Leftrightarrow\left(a^2-\frac{1}{2}\right)^2=\frac{289}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}a^2-\frac{1}{2}=\frac{17}{2}\\a^2-\frac{1}{2}=\frac{-17}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a^2=9\\a^2=-8\left(vôlý\right)́\end{matrix}\right.\)
<=> x=3 hoặc x=-3
