Cho biểu thức: \( B = \left( {{{\sqrt x + 2} \over {\sqrt x – 1}} – {{\sqrt x + 1} \over {\sqrt x – 3}} + {{3\sqrt x – 1} \over {\left( {\sqrt x – 1} \right)\left( {\sqrt x – 3} \right)}}} \right):\left( {1 – {1 \over {\sqrt x – 1}}} \right) \)
a) Xác định tập xác định của biểu thức.
b) Rút gọn biểu thức.
A a) \( \left\{\begin{matrix} x \geq 0\\ x \neq 1 \\ x \neq 9 \end{matrix}\right. \)
b) \( B=\frac{2}{\sqrt{x}-2}\)
B a) \( \left\{\begin{matrix} x \geq 0\\ x \neq 4 \\ x \neq 9 \end{matrix}\right. \)
b) \( B=\frac{2}{\sqrt{x}-2}\)
C a) \( \left\{\begin{matrix} x \geq 0\\ x \neq 1 \\ x \neq 4 \end{matrix}\right. \)
b) \( B=\frac{2}{\sqrt{x}-2}\)
D a) \( \left\{\begin{matrix} x \geq 0\\ x \neq 1 \\ x \neq 4 \\ x \neq 9 \end{matrix}\right. \)
b) \( B=\frac{2}{\sqrt{x}-2}\)
Hướng dẫn Chọn đáp án là: D
Lời giải chi tiết:
a) Hàm số xác định \( \Leftrightarrow \left\{ \matrix{x \ge 0 \hfill \cr \sqrt x – 1 \ne 0 \hfill \cr \sqrt x – 3 \ne 0 \hfill \cr 1 – {1 \over {\sqrt x – 1}} \ne 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{x \ge 0 \hfill \cr \sqrt x \ne 1 \hfill \cr \sqrt x \ne 3 \hfill \cr {1 \over {\sqrt x – 1}} \ne 1 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{x \ge 0 \hfill \cr x \ne 1 \hfill \cr x \ne 9 \hfill \cr \sqrt x – 1 \ne 1 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{x \ge 0 \hfill \cr x \ne 1 \hfill \cr x \ne 9 \hfill \cr \sqrt x \ne 2 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{x \ge 0 \hfill \cr x \ne 1 \hfill \cr x \ne 4 \hfill \cr x \ne 9 \hfill \cr} \right.. \)
\( \eqalign{& b)\,B = \left( {{{\sqrt x + 2} \over {\sqrt x – 1}} – {{\sqrt x + 1} \over {\sqrt x – 3}} + {{3\sqrt x – 1} \over {\left( {\sqrt x – 1} \right)\left( {\sqrt x – 3} \right)}}} \right):\left( {1 – {1 \over {\sqrt x – 1}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\, = {{\left( {\sqrt x + 2} \right)\left( {\sqrt x – 3} \right) – \left( {\sqrt x + 1} \right)\left( {\sqrt x – 1} \right) + 3\sqrt x – 1} \over {\left( {\sqrt x – 1} \right)\left( {\sqrt x – 3} \right)}}:{{\sqrt x – 1 – 1} \over {\sqrt x – 1}} \cr & \,\,\,\,\,\,\,\,\,\,\, = {{x – \sqrt x – 6 – x + 1 + 3\sqrt x – 1} \over {\left( {\sqrt x – 1} \right)\left( {\sqrt x – 3} \right)}}.{{\sqrt x – 1} \over {\sqrt x – 2}} \cr & \,\,\,\,\,\,\,\,\,\,\, = {{2\sqrt x – 6} \over {\left( {\sqrt x – 3} \right)\left( {\sqrt x – 2} \right)}} = {{2\left( {\sqrt x – 3} \right)} \over {\left( {\sqrt x – 3} \right)\left( {\sqrt x – 2} \right)}} = {2 \over {\sqrt x – 2}}. \cr} \)
Chọn D.