Đáp án:
\(\left[ \begin{array}{l}
B = 2\sqrt {a - 1} \\
B = 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ne 4\\
A = \left| {x - 4} \right| + \dfrac{{x - 4}}{{\sqrt {{{\left( {x - 4} \right)}^2}} }} = \left| {x - 4} \right| + \dfrac{{x - 4}}{{\left| {x - 4} \right|}}\\
= \dfrac{{{x^2} - 8x + 16 + x - 4}}{{\left| {x - 4} \right|}}\\
= \dfrac{{{x^2} - 7x + 12}}{{\left| {x - 4} \right|}}\\
\to \left[ \begin{array}{l}
A = \dfrac{{{x^2} - 7x + 12}}{{x - 4}}\left( {DK:x > 4} \right)\\
A = \dfrac{{{x^2} - 7x + 12}}{{ - \left( {x - 4} \right)}}\left( {DK:x < 4} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
A = \dfrac{{\left( {x - 4} \right)\left( {x - 3} \right)}}{{x - 4}} = x - 3\\
A = \dfrac{{\left( {x - 4} \right)\left( {x - 3} \right)}}{{ - \left( {x - 4} \right)}} = - x + 3
\end{array} \right.\\
b.DK:a \ge 1\\
B = \sqrt {a + 2\sqrt {a - 1} } + \sqrt {a - 2\sqrt {a - 1} } \\
= \sqrt {a - 1 + 2\sqrt {a - 1} .1 + 1} + \sqrt {a - 1 - 2\sqrt {a - 1} .1 + 1} \\
= \sqrt {{{\left( {\sqrt {a - 1} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {a - 1} - 1} \right)}^2}} \\
= \sqrt {a - 1} + 1 + \left| {\sqrt {a - 1} - 1} \right|\\
\to \left[ \begin{array}{l}
B = \sqrt {a - 1} + 1 + \sqrt {a - 1} - 1\left( {DK:\sqrt {a - 1} \ge 1 \to a - 1 \ge 1 \to a \ge 2} \right)\\
B = \sqrt {a - 1} + 1 - \sqrt {a - 1} + 1\left( {DK:1 \le a < 2} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
B = 2\sqrt {a - 1} \\
B = 2
\end{array} \right.
\end{array}\)
