Đáp án:

$4) \,\, x = -3$

$5)\,\, x = \dfrac{4}{3}$

$6) \,\, \left[\begin{array}{l}x= \dfrac{41}{2}\\x = \dfrac{5}{2}\end{array}\right.$

Giải thích các bước giải:

$\begin{array}{l}4)\,\,4\sqrt{9-9x} -7\sqrt{1 - x} = 6 + \sqrt{4 - 4x} \,\,\,\,\,\,(ĐK: x \leq 1)\\ \Leftrightarrow 4\sqrt{9(1-x)} -7\sqrt{1 - x} = 6 + \sqrt{4(1 - x)}\\ \Leftrightarrow 12\sqrt{1-x} -7\sqrt{1 - x} = 6 + 2\sqrt{1 - x}\\ \Leftrightarrow 3\sqrt{1 - x} = 6\\ \Leftrightarrow \sqrt{1 - x} = 2\\ \Leftrightarrow 1 - x = 4\\ \Leftrightarrow x = -3 \,\,\,(nhận)\\ \text{Vậy x= -3}\\\\ 5)\,\,3\sqrt{48x} -2\sqrt{12x} = 10 +\sqrt{27x}\,\,\,\,\,\,(ĐK: x \geq 0)\\ \Leftrightarrow 3\sqrt{16.(3x)} -2\sqrt{4.(3x)} = 10 +\sqrt{9.(3x)}\\ \Leftrightarrow 12\sqrt{3x} -4\sqrt{3x} = 10 +3\sqrt{3x}\\ \Leftrightarrow 5\sqrt{3x} = 10\\ \Leftrightarrow \sqrt{3x} = 2\\ \Leftrightarrow 3x = 4\\ \Leftrightarrow x = \dfrac{4}{3}\,\,\,(nhận)\\ \text{Vậy x = $\dfrac{4}{3}$}\\\\ 6)\,\,2\sqrt{2x + 4 - 6\sqrt{2x-5}} - 1 = 5\,\,\,\,\,(ĐK: x \geq \dfrac{5}{2})\\ \Leftrightarrow \sqrt{2x -5 - 6\sqrt{2x-5} + 9} =3\\ \Leftrightarrow \sqrt{(\sqrt{2x - 5} - 3)^2}= 3\\ \Leftrightarrow |\sqrt{2x - 5} - 3| = 3\\ \Leftrightarrow \left[\begin{array}{l}\sqrt{2x - 5} - 3 = 3\,\,\,\,(x \geq 4)\\3 - \sqrt{2x - 5} = 3\,\,\,\,\,(\dfrac{5}{2} \leq x < 4)\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}2x - 5 = 36\\2x - 5 = 0\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x= \dfrac{41}{2}\\x = \dfrac{5}{2}\end{array}\right.\,\,\,\,(nhận)\\ \text{Vậy x = $\dfrac{5}{2}$ hoặc x = $\dfrac{41}{2}$}\end{array}$