hệ pt

Question

Giải phương trình $2(2x^{2}-6x+3)^{2}-6(2x^{2}-6x+3)+3=x$

score 5 · Answer 1

Cách 2:
Đặt: $2x^2-6x+3=y$, ta được hệ:
      $\left\{\begin{array}{l}2x^2-6x+3=y\\2y^2-6y+3=x\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}2x^2-6x+3=y\\2x^2-6x-2y^2+6y=y-x\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}2x^2-6x+3=y\\(x-y)(2x+2y-5)=0\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}2x^2-6x+3=y\\x=y\end{array}\right.&(I)\\\left\{\begin{array}{l}2x^2-6x+3=y\\2x+2y-5=0\end{array}\right.&(II)\end{array}\right.$
$(I)\Leftrightarrow \left\{\begin{array}{l}x=y\\2x^2-6x+3=x\end{array}\right.$
       $\Leftrightarrow \left[\begin{array}{l}x=y=3\\x=y=\frac{1}{2}\end{array}\right.$
$(II)\Leftrightarrow \left\{\begin{array}{l}2y=5-2x\\4x^2-12x+6=5-2x\end{array}\right.$
         $\Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l}x=\frac{5-\sqrt{21}}{4}\\y=\frac{5+\sqrt{21}}{4}\end{array}\right.\\\left\{\begin{array}{l}x=\frac{5+\sqrt{21}}{4}\\y=\frac{5-\sqrt{21}}{4}\end{array}\right.\end{array}\right.$

score 5 · Answer 2

Bình chọn tăng 5 Bình chọn giảm

Cách 1:
Ta có:
$2(2x^2-6x+3)^2-6(2x^2-6x+3)+3=x$
$\Leftrightarrow 2(4x^2-24x^3+48x^2-36x+9)-12x^2+36x-18+3-x=0$
$\Leftrightarrow 8x^4-48x^3+84x^2-37x+3=0$
$\Leftrightarrow (x-3)(8x^3-24x^2+12x-1)=0$
$\Leftrightarrow (x-3)(2x-1)(4x^2-10x+1)=0$
$\Leftrightarrow \left[\begin{array}{l}x=3\\x=\frac{1}{2}\\x=\frac{5\pm\sqrt{21}}{4}\end{array}\right.$