Câu hỏi
Rút gọn \(A = {\left( {\frac{{1 + \sqrt 3 }}{2} - \frac{2}{{1 + \sqrt 3 }}} \right)^2} - {\left( {\frac{{1 - \sqrt 3 }}{2} - \frac{2}{{1 - \sqrt 3 }}} \right)^2}.\)
- A \(A = - 2\sqrt 3 \)
- B \(A = - 3\sqrt 3 \)
- C \(A = 2\sqrt 3 \)
- D \(A = 3\sqrt 3 \)
Phương pháp giải:
Sử dụng hằng đẳng thức \({a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right).\)
Lời giải chi tiết:
\(\begin{array}{l}A = {\left( {\frac{{1 + \sqrt 3 }}{2} - \frac{2}{{1 + \sqrt 3 }}} \right)^2} - {\left( {\frac{{1 - \sqrt 3 }}{2} - \frac{2}{{1 - \sqrt 3 }}} \right)^2}\\A = \left( {\frac{{1 + \sqrt 3 }}{2} - \frac{2}{{1 + \sqrt 3 }} + \frac{{1 - \sqrt 3 }}{2} - \frac{2}{{1 - \sqrt 3 }}} \right)\left( {\frac{{1 + \sqrt 3 }}{2} - \frac{2}{{1 + \sqrt 3 }} - \frac{{1 - \sqrt 3 }}{2} + \frac{2}{{1 - \sqrt 3 }}} \right)\\A = \left[ {\frac{{1 + \sqrt 3 + 1 - \sqrt 3 }}{2} - \left( {\frac{2}{{1 + \sqrt 3 }} + \frac{2}{{1 - \sqrt 3 }}} \right)} \right]\left[ {\frac{{1 + \sqrt 3 - 1 + \sqrt 3 }}{2} + \left( {\frac{2}{{1 - \sqrt 3 }} - \frac{2}{{1 + \sqrt 3 }}} \right)} \right]\\A = \left[ {1 - \frac{{2\left( {1 - \sqrt 3 } \right) + 2\left( {1 + \sqrt 3 } \right)}}{{1 - {{\left( {\sqrt 3 } \right)}^2}}}} \right].\left[ {\sqrt 3 + \frac{{2\left( {1 + \sqrt 3 } \right) - 2\left( {1 - \sqrt 3 } \right)}}{{1 - {{\left( {\sqrt 3 } \right)}^2}}}} \right]\\A = \left( {1 - \frac{{2 - 2\sqrt 3 + 2 + 2\sqrt 3 }}{{ - 2}}} \right)\left( {\sqrt 3 + \frac{{2 + 2\sqrt 3 - 2 + 2\sqrt 3 }}{{ - 2}}} \right)\\A = \left( {1 + \frac{4}{2}} \right)\left( {\sqrt 3 - \frac{{4\sqrt 3 }}{2}} \right) = 3.\frac{{ - 2\sqrt 3 }}{2}\\A = - 3\sqrt 3 .\end{array}\)
Chọn B.