a) \(\left(x^2-2x\right)^2-6\left(x^2-2x\right)+9=0\)
\(\Leftrightarrow\left(x^2-2x\right)^2-2\cdot\left(x^2-2x\right)\cdot3+3^2=0\)
\(\Leftrightarrow\left(x^2-2x-3\right)^2=0\)
\(\Leftrightarrow x^2-2x-3=0\)
\(\Leftrightarrow x^2+x-3x-3=0\)
\(\Leftrightarrow x\left(x+1\right)-3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
Vậy \(S=\left\{-1;3\right\}\).
b) \(\left(x^2+x+1\right)\left(x^2+x+2\right)=12\) (1)
Đặt \(x^2+x+1=y\), khi đó pt (1) trở thành:
\(y\left(y+1\right)=12\)
\(\Leftrightarrow y^2+y-12=0\)
\(\Leftrightarrow y^2+4y-3y-12=0\)
\(\Leftrightarrow y\left(y+4\right)-3\left(y+4\right)=0\)
\(\Leftrightarrow\left(y+4\right)\left(y-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y+4=0\\y-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x^2+x+5=0\\x^2+x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}=0\left(\text{vô lí}\right)\\x^2-x+2x-2=0\end{matrix}\right.\)
\(\Rightarrow x\left(x-1\right)+2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy \(S=\left\{1;-2\right\}\).
c) \(\left(2x^2-3x+1\right)\left(2x^2+5x+1\right)-9x^2=0\) (2)
Đặt \(2x^2+x+1=t\), khi đó pt (2) trở thành:
\(\left(t-4x\right)\left(t+4x\right)-9x^2=0\)
\(\Leftrightarrow t^2-16x^2-9x^2=0\)
\(\Leftrightarrow t^2-25x^2=0\)
\(\Leftrightarrow\left(t-5x\right)\left(t+5x\right)=0\)
\(\Rightarrow\left(2x^2+x+1-5x\right)\left(2x^2+x+1+5x\right)=0\)
\(\Leftrightarrow\left(2x^2-4x+1\right)\left(2x^2+6x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-4x+1=0\\2x^2+6x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2\left(x^2-2x+1\right)-1=0\\2\left(x^2+3x\right)+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2\left(x-1\right)^2=1\\2\left[x^2+2\cdot x\cdot\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2\right]-\dfrac{7}{2}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=\dfrac{1}{2}\\2\left(x+\dfrac{3}{2}\right)^2=\dfrac{7}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x-1=\pm\dfrac{\sqrt{2}}{2}\\x+\dfrac{3}{2}=\pm\dfrac{\sqrt{7}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\pm\sqrt{2}}{2}\\x=\dfrac{-3\pm\sqrt{7}}{2}\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{2\pm\sqrt{2}}{2};\dfrac{-3\pm\sqrt{7}}{2}\right\}\).
\(\text{#}Toru\)