Tập xác định của hàm số \(y=\sqrt{2 x-1}+\sqrt{x+3}\) là
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Lời giải:
Báo saiĐKXĐ: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfKttLearuGlw5gvP1wzaeXatLxBI9gBam % XvP5wqSXMqHnxAJn0BKvguHDwzZbqegm0B1jxALjhiov2DaeHbuLwB % Lnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFf % euY-Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9 % q8aq0-yq-He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqaba % WaaqaafaaakeaadaGabaabaeqabaGaeGOmaiJaemiEaGNaeyOeI0Ia % eGymaeJaeyyzImRaeGimaadabaGaemiEaGNaey4kaSIaeG4mamJaey % yzImRaeGimaadaaiaawUhaaiabgsDiBpaaceaaeaqabeaacqWG4baE % cqGHLjYSdaWcaaqaaiabigdaXaqaaiabikdaYaaaaeaacqWG4baEcq % GHLjYScqGHsislcqaIZaWmaaGaay5EaaGaeyO0H4TaemiEaGNaeyyz % Im7aaSaaaeaacqaIXaqmaeaacqaIYaGmaaaaaa!5E78! \left\{ \begin{array}{l} 2x - 1 \ge 0\\ x + 3 \ge 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ge \frac{1}{2}\\ x \ge - 3 \end{array} \right. \Rightarrow x \ge \frac{1}{2}\)
Vậy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfKttLearuGlw5gvP1wzaeXatLxBI9gBam % XvP5wqSXMqHnxAJn0BKvguHDwzZbqegm0B1jxALjhiov2DaeHbuLwB % Lnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFf % euY-Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9 % q8aq0-yq-He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqaba % WaaqaafaaakeaacqWGebarcqGH9aqpdaqcsaqaamaalaaabaGaeGym % aedabaGaeGOmaidaaiabcUda7iabgUcaRiabg6HiLcGaay5waiaawM % caaaaa!4429! D = \left[ {\frac{1}{2}; + \infty } \right)\) hay \(D=\mathbb{R} \backslash\left(-\infty, \frac{1}{2}\right)\)