Đáp án:
m. \(2\sqrt {x - 1} \)
Giải thích các bước giải:
\(\begin{array}{l}
e.DK:x \ne 4;x \ge 0\\
E = \dfrac{{{{\left( {\sqrt x - 2} \right)}^2}}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}} = \dfrac{{\sqrt x - 2}}{{\sqrt x + 2}}\\
g.DK:x \ge 0;x \ne 9\\
G = \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 3} \right)}}{{\sqrt x - 3}}\\
= \sqrt x - 1\\
h.H = \sqrt {{{\left( {2x - 1} \right)}^2}} - \sqrt {{{\left( {2x + 1} \right)}^2}} \\
= \left| {2x - 1} \right| - \left| {2x + 1} \right|\\
\to \left[ \begin{array}{l}
H = 2x - 1 - 2x - 1\\
H = 2x - 1 + 2x + 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
H = - 2\\
H = 4x
\end{array} \right.\\
i.I = \left| {2x - \sqrt {{{\left( {5x - 1} \right)}^2}} } \right|\\
= \left( { - 2x} \right) - \left| {5x - 1} \right|\\
= - 2x - \left( { - 5x + 1} \right)\\
= - 2x + 5x - 1 = 3x - 1\left( {Do:x < 0} \right)\\
k.K = \dfrac{2}{{x - 1}}.\dfrac{{\sqrt {{{\left( {x - 1} \right)}^2}} }}{{2x}}\left( {do:0 < x < 1} \right)\\
= \dfrac{2}{{x - 1}}.\dfrac{{ - \left( {x - 1} \right)}}{{2x}}\\
= - \dfrac{1}{x}\\
m.DK:x \ge 1\\
M = \sqrt {x - 1 + 2\sqrt {x - 1} + 1} + \sqrt {x - 1 - 2\sqrt {x - 1} + 1} \\
= \sqrt {{{\left( {\sqrt {x - 1} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {x - 1} - 1} \right)}^2}} \\
= \left| {\sqrt {x - 1} + 1} \right| + \left| {\sqrt {x - 1} - 1} \right|\\
= \sqrt {x - 1} + 1 + \sqrt {x - 1} - 1\\
= 2\sqrt {x - 1}
\end{array}\)
