Giải thích các bước giải:

a.

$
\dfrac{3}{5}(1-5x) - \dfrac{5}{4}(2x-8) = 5x + \dfrac{17}{2} \\
\dfrac{3}{5} - 3x - \dfrac{5}{2}x + 10 = 5x + \dfrac{17}{2} \\
\dfrac{3}{5} + 10 - \dfrac{17}{2} = 5x + 3x + \dfrac{5}{2}x \\
\dfrac{6 + 100 - 85}{10} = \dfrac{10x + 6x + 5x}{2} \\
\dfrac{21}{10} = \dfrac{21x}{2} \\
\dfrac{21}{10} = \dfrac{105x}{10} \\
105x = 21 \\
x = \dfrac{21}{105} \\
x = \dfrac{1}{5}
$

b. 

$
(4x-7)^2 - 5|7-4x| = 0 \\
|4x-7|^2 - 5|4x-7| = 0 \\
|4x-7|(|4x-7| - 5) = 0 \\
$

$\left[ \begin{array}{l} |4x-7| = 0 \\ |4x-7| - 5 = 0 \end{array} \right.
\iff
\left[ \begin{array}{l} 4x-7 = 0 \\ |4x-7| = 5 \end{array} \right.
\iff
\left[ \begin{array}{l} x = \dfrac{7}{4} \\ 4x-7 = 5 \\ 4x-7 = -5 \end{array} \right.$

$\iff
\left[ \begin{array}{l} x = \dfrac{7}{4} \\ 4x = 12 \\ 4x = 2 \end{array} \right.
\iff
\left[ \begin{array}{l} x = 1,75 \\ x = 3 \\ x = 0,5 \end{array} \right.$

c.ĐK: $x - 1 \ge 0 \iff x \ge 1$

$
\sqrt{16(x-1)} - \sqrt{9x-9} = 5 \\
\sqrt{16}\cdot\sqrt{x-1} - \sqrt{9(x-1)} = 5 \\
4\sqrt{x-1} - 3\sqrt{x-1} = 5 \\
(4-3)\sqrt{x-1} = 5 \\
\sqrt{x-1} = 5 \\
x - 1 = 25 \\
x = 26 \quad (\text{Thỏa mãn})
$

d.

$
4^{x-2} + 4^{x+1} = 1040 \\
4^{x-2} + 4^{x-2} \cdot 4^3 = 1040 \\
4^{x-2}(1 + 4^3) = 1040 \\
4^{x-2}(1 + 64) = 1040 \\
4^{x-2} \cdot 65 = 1040 \\
4^{x-2} = \dfrac{1040}{65} \\
4^{x-2} = 16 \\
4^{x-2} = 4^2 \\
x - 2 = 2 \\
x = 4
$