Đáp án:
$\begin{array}{l}
70)\\
Đặt:\sqrt {6{x^2} - 12x + 7} = t\left( {t \ge 0} \right)\\
\Leftrightarrow 6{x^2} - 12x + 7 = {t^2}\\
\Leftrightarrow 2x - {x^2} = \frac{{7 - {t^2}}}{6}\\
Khi:2x - {x^2} + \sqrt {6{x^2} - 12x + 7} = 0\\
\Leftrightarrow \frac{{7 - {t^2}}}{6} + t = 0\\
\Leftrightarrow 7 - {t^2} + 6t = 0\\
\Leftrightarrow {t^2} - 6t - 7 = 0\\
\Leftrightarrow \left( {t - 7} \right)\left( {t + 1} \right) = 0\\
\Leftrightarrow t = 7\left( {do:t \ge 0} \right)\\
\Leftrightarrow 2x - {x^2} = \frac{{7 - {t^2}}}{6} = - 7\\
\Leftrightarrow {x^2} - 2x - 7 = 0\\
\Leftrightarrow {x^2} - 2x + 1 = 8\\
\Leftrightarrow {\left( {x - 1} \right)^2} = 8\\
\Leftrightarrow x = 1 \pm 2\sqrt 2 \\
Vậy\,x = 1 \pm 2\sqrt 2 \\
71)Dkxd:{x^2} \ge 7 \Leftrightarrow \left[ \begin{array}{l}
x \ge \sqrt 7 \\
x \le - \sqrt 7
\end{array} \right.\\
\sqrt {{x^2} + 9} - \sqrt {{x^2} - 7} = 2\\
\Leftrightarrow \sqrt {{x^2} + 9} = \sqrt {{x^2} - 7} + 2\\
\Leftrightarrow {x^2} + 9 = {x^2} - 7 + 4\sqrt {{x^2} - 7} + 4\\
\Leftrightarrow 4\sqrt {{x^2} - 7} = 12\\
\Leftrightarrow \sqrt {{x^2} - 7} = 3\\
\Leftrightarrow {x^2} - 7 = 9\\
\Leftrightarrow {x^2} = 16\\
\Leftrightarrow x = 4;x = - 4\left( {tmdk} \right)\\
Vậy\,x = 4;x = - 4\\
72)Dkxd:x \ge 4\\
\sqrt {x + 3} - \sqrt {x - 4} = 1\\
\Leftrightarrow \sqrt {x + 3} = \sqrt {x - 4} + 1\\
\Leftrightarrow x + 3 = x - 4 + 2\sqrt {x - 4} + 1\\
\Leftrightarrow 2\sqrt {x - 4} = 6\\
\Leftrightarrow \sqrt {x - 4} = 3\\
\Leftrightarrow x - 4 = 9\\
\Leftrightarrow x = 13\left( {tmdk} \right)\\
Vậy\,x = 13
\end{array}$
