$\eqalign{
& \sqrt {1 - x} + \sqrt {1 + x} = 2 - {{{x^2}} \over 4}\;(1) \cr
& dk: - 1 \le x \le 1 \cr
& (1) \leftrightarrow \sqrt {1 - x} - 1 + \sqrt {1 + x} - 1 = - {{{x^2}} \over 4} \cr
& \leftrightarrow {{1 - x - 1} \over {\sqrt {1 - x} + 1}} + {{1 + x - 1} \over {\sqrt {1 + x} + 1}} = - {{{x^2}} \over 4} \cr
& \leftrightarrow x\left( {{1 \over {\sqrt {1 + x} + 1}} - {1 \over {\sqrt {1 - x} + 1}} + {x \over 4}} \right) = 0 \cr
& TH1:x = 0 \cr
& TH2:{1 \over {\sqrt {1 + x} + 1}} - {1 \over {\sqrt {1 - x} + 1}} + {x \over 4} = 0 \cr
& \leftrightarrow {{\sqrt {1 - x} + 1 - \sqrt {1 + x} - 1} \over {\left( {\sqrt {1 + x} + 1} \right)\left( {\sqrt {1 - x} + 1} \right)}} + {x \over 4} = 0\;(quy\;dong) \cr
& \leftrightarrow {{1 - x - 1 - x} \over {\left( {\sqrt {1 + x} + 1} \right)\left( {\sqrt {1 - x} + 1} \right)\left( {\sqrt {1 - x} + \sqrt {1 + x} } \right)}} + {x \over 4} = 0 \leftrightarrow \cr
& - x\left( {{2 \over {\left( {\sqrt {1 + x} + 1} \right)\left( {\sqrt {1 - x} + 1} \right)\left( {\sqrt {1 - x} + \sqrt {1 + x} } \right)}} - {1 \over 4}} \right) = 0 \cr
& \leftrightarrow x = 0 \cr
& do: \cr
& bunhiacopski: \cr
& \sqrt {1 - x} + \sqrt {1 + x} = 1*\sqrt {1 - x} + 1*\sqrt {1 + x} \le \sqrt {\left( {1 + 1} \right)\left( {1 - x + 1 + x} \right)} = 2 \cr
& ma: \cr
& \left( {\sqrt {1 + x} + 1} \right)\left( {\sqrt {1 - x} + 1} \right) = \sqrt {\left( {1 + x} \right)\left( {1 - x} \right)} + \sqrt {1 - x} + \sqrt {1 + x} + 1 \le \sqrt {\left( {1 + x} \right)\left( {1 - x} \right)} + 3\;\;(\sqrt {1 - x} + \sqrt {1 + x} \le 2) \cr
& cosi: \cr
& \sqrt {\left( {1 + x} \right)\left( {1 - x} \right)} \le {{1 + x} \over 2} + {{1 - x} \over 2} = 1\;(do\;dk\; - 1 \le x \le 1) \cr
& \to \left( {\sqrt {1 + x} + 1} \right)\left( {\sqrt {1 - x} + 1} \right) \le 4 \cr
& \to \left( {\sqrt {1 + x} + 1} \right)\left( {\sqrt {1 - x} + 1} \right)\left( {\sqrt {1 - x} + \sqrt {1 + x} } \right) \le 8 \cr
& \leftrightarrow {8 \over {\left( {\sqrt {1 + x} + 1} \right)\left( {\sqrt {1 - x} + 1} \right)\left( {\sqrt {1 - x} + \sqrt {1 + x} } \right)}} \ge 1 \cr
& \leftrightarrow {2 \over {\left( {\sqrt {1 + x} + 1} \right)\left( {\sqrt {1 - x} + 1} \right)\left( {\sqrt {1 - x} + \sqrt {1 + x} } \right)}} - {1 \over 4} \ge 0 \cr
& dau\; = \leftrightarrow x = 0 \cr} $
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
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