Đáp án:
b. \(\left[ \begin{array}{l}
B = \dfrac{1}{{\sqrt x - 1}} + \dfrac{2}{{x - 2}}\\
B = \dfrac{1}{{\sqrt x - 1}} + \dfrac{2}{{ - x + 2}}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:3 \le x < 4\\
A = \sqrt {x - 3} - \sqrt {\dfrac{1}{{4 - x}}} \\
= \dfrac{{\sqrt {\left( {x - 3} \right)\left( {4 - x} \right)} - 1}}{{\sqrt {4 - x} }}\\
= \dfrac{{\sqrt { - {x^2} + 7x - 12} - 1}}{{\sqrt {4 - x} }}\\
b.DK:x \ge 0;x \ne \left\{ {1;2} \right\}\\
B = \dfrac{1}{{\sqrt x - 1}} + \dfrac{2}{{\sqrt {{x^2} - 4x + 4} }}\\
= \dfrac{1}{{\sqrt x - 1}} + \dfrac{2}{{\sqrt {{{\left( {x - 2} \right)}^2}} }}\\
= \dfrac{1}{{\sqrt x - 1}} + \dfrac{2}{{\left| {x - 2} \right|}}\\
\to \left[ \begin{array}{l}
B = \dfrac{1}{{\sqrt x - 1}} + \dfrac{2}{{x - 2}}\left( {DK:x > 2} \right)\\
B = \dfrac{1}{{\sqrt x - 1}} + \dfrac{2}{{ - x + 2}}\left( {DK:0 \le x < 2;x \ne 1} \right)
\end{array} \right.
\end{array}\)
