Cho A = 1 / (2 + 2 căn bậc hai a) + 1 / (2 - 2 căn bậc hai a) - (a^2 + 1) / (1 - a^2)
Câu hỏi:
Cho \(A = \frac{1}{{2 + 2\sqrt a }} + \frac{1}{{2 - 2\sqrt a }} - \frac{{{a^2} + 1}}{{1 - {a^2}}}\)
a) Tìm điều kiện xác định rồi rút gọn A
b) Tìm a để \[{\rm{A}} < \frac{1}{3}\].
Trả lời:
a) Điều kiện xác định a ≥ 0, a ≠ 1
\(A = \frac{1}{{2 + 2\sqrt a }} + \frac{1}{{2 - 2\sqrt a }} - \frac{{{a^2} + 1}}{{1 - {a^2}}}\)
\({\rm{A}} = \frac{1}{{2\left( {1 + \sqrt a } \right)}} + \frac{1}{{2\left( {1 - \sqrt a } \right)}} - \frac{{{a^2} + 1}}{{\left( {1 - a} \right)\left( {1 + a} \right)}}\)
\({\rm{A}} = \frac{1}{{2\left( {1 + \sqrt a } \right)}} + \frac{1}{{2\left( {1 - \sqrt a } \right)}} - \frac{{{a^2} + 1}}{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a } \right)\left( {1 + a} \right)}}\)
\(A = \frac{{\left( {1 - \sqrt a } \right)\left( {1 + a} \right) + \left( {1 + \sqrt a } \right)\left( {1 + a} \right) - \left( {{a^2} + 1} \right)2}}{{2\left( {1 + \sqrt a } \right)\left( {1 - \sqrt a } \right)\left( {1 + a} \right)}}\)
\(A = \frac{{1 + a - \sqrt a - a\sqrt a + 1 + a + \sqrt a + a\sqrt a - 2{a^2} + 2}}{{2\left( {1 + \sqrt a } \right)\left( {1 - \sqrt a } \right)\left( {1 + a} \right)}}\)
\({\rm{A}} = \frac{{2a - 2{a^2}}}{{2\left( {1 - a} \right)\left( {1 + a} \right)}}\)
\({\rm{A}} = \frac{{2a\left( {1 - a} \right)}}{{2\left( {1 - a} \right)\left( {1 + a} \right)}}\)
\({\rm{A}} = \frac{a}{{1 + a}}\)
b) Để \[{\rm{A}} < \frac{1}{3}\]\( \Leftrightarrow \frac{a}{{1 + a}} < \frac{1}{3}\)
\( \Leftrightarrow \frac{a}{{1 + a}} - \frac{1}{3} < 0\)\( \Leftrightarrow \frac{{3a - a - 1}}{{1 + a}} < 0\)
\( \Leftrightarrow 2{\rm{a}} - 1 < 0\)\( \Leftrightarrow {\rm{a}} < \frac{1}{2}\)
Mà a ≥ 0, a ≠ 1
Suy ra \({\rm{0}} \le {\rm{a}} < \frac{1}{2}\)
Vậy \({\rm{0}} \le {\rm{a}} < \frac{1}{2}\).
