Đáp án:
$c) \, x = \left\{\pm \sqrt{5}; \, \pm 2\right\}$
$d) \, x = \left\{\dfrac{1}{3}\right\}$
Giải thích các bước giải:
c) $\sqrt{x^2 - 4} - x^2 + 4 = 0 \, (*)$
$ĐK: \, x^2 - 4 \geq 0 \Leftrightarrow \left[\begin{array}{l}x\geq 2\\x \leq -2 \end{array}\right.$
$(*) \Leftrightarrow \sqrt{x^2 - 4} - (x^2 - 4) = 0$
$\Leftrightarrow \sqrt{x^2 - 4}(1 - \sqrt{x^2 - 4}) = 0$
$\Leftrightarrow \left[\begin{array}{l}\sqrt{x^2 - 4} = 0\\\sqrt{x^2 - 4} = 1\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = \pm 2\\x = \pm \sqrt{5}\end{array}\right. \,\,\,\, (nhận)$
Vậy $S= \left\{\pm \sqrt{5}; \, \pm 2\right\}$
d) $\sqrt{9x^2 - 6x + 2} + \sqrt{45x^2 - 30x + 9} = \sqrt{6x - 9x^2 + 8}$ $(**)$
$ĐK: \, 6x - 9x^2 + 8 \geq 0 \Leftrightarrow (3x - 1)^2 \leq 9 \Leftrightarrow -3 \leq 3x - 1 \leq 3 \Leftrightarrow -\dfrac{2}{3} \leq x \leq \dfrac{4}{3}$
$(**) \Leftrightarrow \sqrt{(3x -1)^2 + 1} + \sqrt{5(3x -1)^2 + 4} = \sqrt{9 - (3x -1)^2}$
Đặt $t = (3x -1)^2$, ta được:
$\sqrt{t +1} + \sqrt{5t + 4} = \sqrt{9 - t}$
$\Leftrightarrow t + 1 + 5t + 4 + 2\sqrt{(t+1)(5t +4)} = 9 -t$
$\Leftrightarrow 2\sqrt{(t +1)(5t +4)} = 4 - 7t$
$\Leftrightarrow 4(5t^2 + 9t + 4) = 16 - 56t+ 49t^2$
$\Leftrightarrow 29t^2 - 92t = 0$
$\Leftrightarrow t(29t - 92) = 0$
$\Leftrightarrow \left[\begin{array}{l}t = 0\\t = \dfrac{92}{29} \, \, (loại)\end{array}\right.$
$\Leftrightarrow (3x-1)^2 = 0$
$\Leftrightarrow x = \dfrac{1}{3} \, \, (thoả \,\, ĐK)$
Vậy $ x = \dfrac{1}{3}$
