PT \(\Leftrightarrow\left(x^2-3x\right)^2+2\left(x^2-3x\right)+3\left(x^2-3x\right)+6=0\)
\(\Leftrightarrow\left(x^2-3x\right)\left(x^2-3x+2\right)+3\left(x^2-3x+2\right)=0\)
\(\Leftrightarrow\left(x^2-3x+2\right)\left(x^2-3x+3\right)=0\)
Thấy : \(x^2-2.x.\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{3}{4}=\left(x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)
\(\Rightarrow x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy ...
Đặt: x2 - 3x = t thì PT trở thành: t2 + 5t + 6 = 0
⇔ (t + 2)(t + 3) = 0
\(\Leftrightarrow\left[{}\begin{matrix}t=-2\\t=-3\end{matrix}\right.\)
Với t = -2 ⇒ x2 - 3x + 2 = 0 ⇔ (x - 2)(x - 1) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
Với t = -3 ⇒ x2 - 3x + 3 = 0
\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}=0\) (vô lý vì: \(\left(x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\))
Vậy...
Bạn tham khảo nhé!