$4\sqrt{2x - 3} = - x^2 + 12x - 15$
$ĐK: \, \begin{cases} 2x - 3 \geq 0\\-x^2 + 12x - 15 \geq 0\end{cases}$
$\Leftrightarrow \begin{cases} x \geq \dfrac{3}{2}\\-(x - 6)^2 + 21 \geq 0 \end{cases}$
$\Leftrightarrow \begin{cases} x \geq \dfrac{3}{2}\\(x-6)^2 \leq 21 \end{cases}$
$\Leftrightarrow \begin{cases} x \geq \dfrac{3}{2}\\-\sqrt{21} + 6 \leq x \leq \sqrt{21} + 6 \end{cases}$
$\Leftrightarrow \dfrac{3}{2} \leq x \leq \sqrt{21} + 6$
Ta có:
$4\sqrt{2x - 3} = - x^2 + 12x - 15$
$\Leftrightarrow \sqrt{16(2x - 3)} = -(x - 6)^2 + 21$
$\Leftrightarrow 32x - 48 = [-(x-6)^2 + 21]^2$
$\Leftrightarrow 32(x - 6) + 144 = (x - 6)^4 - 42(x - 6)^2 + 441$
$\Leftrightarrow (x-6)^4 - 42(x - 6)^2 - 32(x - 6) + 297 = 0$
Đặt $x - 6 = t$, ta được:
$t^4 - 42t^2 - 32t + 297 = 0$
$\Leftrightarrow t^4 - 4t^2 - 11t^2 - 27t^2 + 22t - 54t + 27.11 = 0$
$\Leftrightarrow (t^4 - 4t^2) - (11t^2 - 22t) - (27t^2 + 54t) + 27.11 = 0$
$\Leftrightarrow (t^2 - 2t)(t^2 + 2t) - 11(t^2 - 2t) - 27(t^2 + 2t) + 27.11 = 0$
$\Leftrightarrow (t^2 - 2t)(t^2 + 2t - 11) - 27(t^2 + 2t - 11) = 0$
$\Leftrightarrow (t^2 - 2t - 27)(t^2 + 2t - 11) = 0$
$\Leftrightarrow$ \(\left[ \begin{array}{l}t=-1-2\sqrt{3}\\t=-1+2\sqrt{3}\\t=1+2\sqrt{7}\\t=1 - 2\sqrt{7}\end{array} \right.\)
$\Rightarrow$ \(\left[ \begin{array}{l}x=5-2\sqrt{3}\\x=5+2\sqrt{3}\\x=7+2\sqrt{7} \, (loại)\\x=7 - 2\sqrt{7}\end{array} \right.\)
