Câu hỏi
Rút gọn các biểu thức sau:
Câu 1: \(P = \left( {\frac{2}{{\sqrt x - 2}} + \frac{{\sqrt x - 1}}{{2\sqrt x - x}}} \right):\left( {\frac{{\sqrt x + 2}}{{\sqrt x }} - \frac{{\sqrt x - 1}}{{\sqrt x - 2}}} \right)\)
- A \(P= \frac{{\sqrt x + 1}}{{\sqrt x - 4}}\)
- B \(P= \frac{{\sqrt x - 1}}{{\sqrt x - 4}}\)
- C \(P= \frac{{\sqrt x + 1}}{{\sqrt x + 4}}\)
- D \(P= \frac{{\sqrt x - 1}}{{\sqrt x + 4}}\)
Lời giải chi tiết:
\(P = \left( {\frac{2}{{\sqrt x - 2}} + \frac{{\sqrt x - 1}}{{2\sqrt x - x}}} \right):\left( {\frac{{\sqrt x + 2}}{{\sqrt x }} - \frac{{\sqrt x - 1}}{{\sqrt x - 2}}} \right)\)
Điều kiện: \(x > 0;x \ne 4\)
\(\begin{array}{l}P = \left( {\frac{2}{{\sqrt x - 2}} + \frac{{\sqrt x - 1}}{{2\sqrt x - x}}} \right):\left( {\frac{{\sqrt x + 2}}{{\sqrt x }} - \frac{{\sqrt x - 1}}{{\sqrt x - 2}}} \right)\\\,\,\,\, = \left( {\frac{2}{{\sqrt x - 2}} - \frac{{\sqrt x - 1}}{{\sqrt x \left( {\sqrt x - 2} \right)}}} \right):\left( {\frac{{\sqrt x + 2}}{{\sqrt x }} - \frac{{\sqrt x - 1}}{{\sqrt x - 2}}} \right)\\\,\,\, = \frac{{\sqrt x + 1}}{{\sqrt x \left( {\sqrt x - 2} \right)}}:\frac{{x - 4 - x + \sqrt x }}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\\,\,\, = \frac{{\sqrt x + 1}}{{\sqrt x \left( {\sqrt x - 2} \right)}}.\frac{{\sqrt x \left( {\sqrt x - 2} \right)}}{{\sqrt x - 4}}\\\,\, = \frac{{\sqrt x + 1}}{{\sqrt x - 4}}.\end{array}\)
Câu 2: \(Q = \left( {\frac{{x + 2}}{{\sqrt x + 1}} - \sqrt x } \right):\left( {\frac{{\sqrt x - 4}}{{1 - x}} - \frac{{\sqrt x }}{{\sqrt x + 1}}} \right)\)
- A \(Q= \frac{{\sqrt x + 1}}{{\sqrt x - 2}}\)
- B \(Q= \frac{{\sqrt x - 1}}{{\sqrt x - 2}}\)
- C \(Q= \frac{{\sqrt x + 1}}{{\sqrt x + 2}}\)
- D \(Q= \frac{{\sqrt x - 1}}{{\sqrt x + 2}}\)
Lời giải chi tiết:
\(Q = \left( {\frac{{x + 2}}{{\sqrt x + 1}} - \sqrt x } \right):\left( {\frac{{\sqrt x - 4}}{{1 - x}} - \frac{{\sqrt x }}{{\sqrt x + 1}}} \right)\)
Điều kiện: \(x \ge 0,\,\,x \ne 1,\,\,x \ne 4.\)
\(\begin{array}{l}Q = \left( {\frac{{x + 2}}{{\sqrt x + 1}} - \sqrt x } \right):\left( {\frac{{\sqrt x - 4}}{{1 - x}} - \frac{{\sqrt x }}{{\sqrt x + 1}}} \right)\\\,\,\,\,\, = \left( {\frac{{x + 2 - \sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}}} \right):\left( {\frac{{4 - \sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}} - \frac{{\sqrt x }}{{\sqrt x + 1}}} \right)\\\,\,\,\, = \frac{{x + 2 - x - \sqrt x }}{{\sqrt x + 1}}:\frac{{4 - \sqrt x - \sqrt x \left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\\,\,\,\, = \frac{{2 - \sqrt x }}{{\sqrt x + 1}}:\frac{{4 - \sqrt x - x + \sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}} = \frac{{2 - \sqrt x }}{{\sqrt x + 1}}:\frac{{4 - x}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\\,\,\, = \frac{{2 - \sqrt x }}{{\sqrt x + 1}}.\frac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}{{\left( {2 - \sqrt x } \right)\left( {2 + \sqrt x } \right)}}\, = \frac{{\sqrt x - 1}}{{\sqrt x + 2}}.\end{array}\)
Câu 3: \(R = \left( {\frac{{3\sqrt x }}{{\sqrt x + 2}} + \frac{{\sqrt x }}{{\sqrt x - 2}} + \frac{{3x - 5\sqrt x }}{{4 - x}}} \right):\left( {\frac{{2\sqrt x - 1}}{{\sqrt x - 2}} - 1} \right)\)
- A \(R= -\frac{{\sqrt x }}{{\sqrt x - 2}}\)
- B \(R= \frac{{\sqrt x }}{{\sqrt x - 2}}\)
- C \(R= \frac{{\sqrt x }}{{\sqrt x + 2}}\)
- D \(R= -\frac{{\sqrt x }}{{\sqrt x + 2}}\)
Lời giải chi tiết:
\(R = \left( {\frac{{3\sqrt x }}{{\sqrt x + 2}} + \frac{{\sqrt x }}{{\sqrt x - 2}} + \frac{{3x - 5\sqrt x }}{{4 - x}}} \right):\left( {\frac{{2\sqrt x - 1}}{{\sqrt x - 2}} - 1} \right)\)
Điều kiện: \(x \ge 0,\,\,\,x \ne 4.\)
\(\begin{array}{l}R = \left( {\frac{{3\sqrt x }}{{\sqrt x + 2}} + \frac{{\sqrt x }}{{\sqrt x - 2}} + \frac{{3x - 5\sqrt x }}{{4 - x}}} \right):\left( {\frac{{2\sqrt x - 1}}{{\sqrt x - 2}} - 1} \right)\\\,\,\,\, = \left( {\frac{{3\sqrt x }}{{\sqrt x + 2}} + \frac{{\sqrt x }}{{\sqrt x - 2}} - \frac{{3x - 5\sqrt x }}{{(\sqrt x - 2)(\sqrt x + 2)}}} \right):\left( {\frac{{2\sqrt x - 1 - \sqrt x + 2}}{{\sqrt x - 2}}} \right)\\\,\,\,\, = \frac{{3\sqrt x \left( {\sqrt x - 2} \right) + \sqrt x \left( {\sqrt x + 2} \right) - 3x + 5\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}:\frac{{\sqrt x + 1}}{{\sqrt x - 2}}\\\,\,\,\, = \frac{{3x - 6\sqrt x + x + 2\sqrt x - 3x + 5\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}:\frac{{\sqrt x + 1}}{{\sqrt x - 2}}\\\,\,\, = \frac{{x + \sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}:\frac{{\sqrt x + 1}}{{\sqrt x - 2}}\\\,\,\, = \frac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\frac{{\sqrt x - 2}}{{\sqrt x + 1}}\\\,\,\, = \frac{{\sqrt x }}{{\sqrt x + 2}}.\end{array}\)
Câu 4: \(S = \left( {\frac{{\sqrt x + 3}}{{\sqrt x - 2}} + \frac{{\sqrt x + 2}}{{3 - \sqrt x }} + \frac{{\sqrt x + 2}}{{x - 5\sqrt x + 6}}} \right):\left( {1 - \frac{{\sqrt x }}{{\sqrt x + 1}}} \right)\)
- A \(S= \frac{{\sqrt x - 1}}{{\sqrt x - 2}}\)
- B \(S= \frac{{\sqrt x + 1}}{{\sqrt x - 2}}\)
- C \(S= \frac{{\sqrt x + 1}}{{\sqrt x + 2}}\)
- D \(S= \frac{{\sqrt x - 1}}{{\sqrt x + 2}}\)
Lời giải chi tiết:
\(\begin{array}{l}4)\,\,S = \left( {\frac{{\sqrt x + 3}}{{\sqrt x - 2}} + \frac{{\sqrt x + 2}}{{3 - \sqrt x }} + \frac{{\sqrt x + 2}}{{x - 5\sqrt x + 6}}} \right):\left( {1 - \frac{{\sqrt x }}{{\sqrt x + 1}}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {DK:\,\,x \ge 0,\,\,x \ne 4,\,\,x \ne 9} \right)\\\,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{{\sqrt x + 3}}{{\sqrt x - 2}} - \frac{{\sqrt x + 2}}{{\sqrt x - 3}} + \frac{{\sqrt x + 2}}{{(\sqrt x - 3).(\sqrt x - 2)}}} \right):\left( {\frac{{\sqrt x + 1 - \sqrt x }}{{\sqrt x + 1}}} \right)\\\,\,\,\,\,\,\,\,\,\, = \frac{{(\sqrt x + 3)(\sqrt x - 3) - (\sqrt x + 2)(\sqrt x - 2) + \sqrt x + 2}}{{(\sqrt x - 3).(\sqrt x - 2)}}:\frac{1}{{\sqrt x + 1}}\\\,\,\,\,\,\,\,\,\,\, = \frac{{x - 9 - (x - 4) + \sqrt x + 2}}{{(\sqrt x - 3).(\sqrt x - 2)}}:\frac{1}{{\sqrt x + 1}}\\\,\,\,\,\,\,\,\,\, = \frac{{x - 9 - x + 4 + \sqrt x + 2}}{{(\sqrt x - 3).(\sqrt x - 2)}}:\frac{1}{{\sqrt x + 1}}\\\,\,\,\,\,\,\,\, = \frac{{\sqrt x - 3}}{{(\sqrt x - 3).(\sqrt x - 2)}}.(\sqrt x + 1)\\\,\,\,\,\,\,\,\, = \frac{{\sqrt x + 1}}{{\sqrt x - 2}}.\end{array}\)
Câu 5: \(T = \frac{{\sqrt x + 1}}{{x - 1}} - \frac{{x + 2}}{{x\sqrt x - 1}} - \frac{{\sqrt x + 1}}{{x + \sqrt x + 1}}\)
- A \(T= \frac{{ - \sqrt x }}{{x + \sqrt x + 1}}\)
- B \(T= \frac{{ - \sqrt x }}{{x - \sqrt x + 1}}\)
- C \(T= \frac{{ \sqrt x }}{{x + \sqrt x + 1}}\)
- D \(T= \frac{{ -2 \sqrt x }}{{x + \sqrt x + 1}}\)
Lời giải chi tiết:
\(\begin{array}{l}\,\,\,T = \frac{{\sqrt x + 1}}{{x - 1}} - \frac{{x + 2}}{{x\sqrt x - 1}} - \frac{{\sqrt x + 1}}{{x + \sqrt x + 1}}\,\,\,\,\,\left( {DK:\,\,x \ge 0,\,\,x \ne 1} \right)\\\,\,\,\,\,\,\,\,\,\,\, = \frac{{\sqrt x + 1}}{{(\sqrt x - 1)(\sqrt x + 1)}} - \frac{{x + 2}}{{(\sqrt x - 1)(x + \sqrt x + 1)}} - \frac{{\sqrt x + 1}}{{x + \sqrt x + 1}}\\\,\,\,\,\,\,\,\,\,\, = \frac{1}{{\sqrt x - 1}} - \frac{{x + 2}}{{(\sqrt x - 1)(x + \sqrt x + 1)}} - \frac{{\sqrt x + 1}}{{x + \sqrt x + 1}}\\\,\,\,\,\,\,\,\,\,\, = \frac{{x + \sqrt x + 1 - (x + 2) - (\sqrt x + 1)(\sqrt x - 1)}}{{(\sqrt x - 1)(x + \sqrt x + 1)}}\\\,\,\,\,\,\,\,\,\,\, = \frac{{x + \sqrt x + 1 - x - 2 - (x - 1)}}{{(\sqrt x - 1)(x + \sqrt x + 1)}}\\\,\,\,\,\,\,\,\,\, = \frac{{\sqrt x - 1 - x + 1}}{{(\sqrt x - 1)(x + \sqrt x + 1)}}\\\,\,\,\,\,\,\,\, = \frac{{\sqrt x - x}}{{(\sqrt x - 1)(x + \sqrt x + 1)}}\\\,\,\,\,\,\,\,\, = \frac{{ - \sqrt x \left( {\sqrt x - 1} \right)}}{{(\sqrt x - 1)(x + \sqrt x + 1)}}\\\,\,\,\,\,\,\,\, = \frac{{ - \sqrt x }}{{x + \sqrt x + 1}}.\end{array}\)
Câu hỏi trước
