Cho hàm số \(f(x)=\left\{\begin{array}{ll} \frac{\sqrt{x+2}-3}{x+1} & x \geq 2 \\ x^{2}-3 x+1 & x<2 \end{array}\right.\). Giá trị của biểu thức\(f(-2)+5 f(4)\) bằng bao nhiêu?
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Lời giải:
Báo saiTa có
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfKttLearuGlw5gvP1wzaeXatLxBI9gBam % XvP5wqSXMqHnxAJn0BKvguHDwzZbqegm0B1jxALjhiov2DaeHbuLwB % Lnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFf % euY-Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9 % q8aq0-yq-He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqaba % WaaqaafaaakqaabeqaaiabgkHiTiabikdaYiabgYda8iabikdaYiab % gkDiElabdAgaMnaabmaabaGaeyOeI0IaeGOmaidacaGLOaGaayzkaa % Gaeyypa0ZaaeWaaeaacqGHsislcqaIYaGmaiaawIcacaGLPaaadaah % aaWcbeqaaiabikdaYaaakiabgkHiTiabiodaZmaabmaabaGaeyOeI0 % IaeGOmaidacaGLOaGaayzkaaGaey4kaSIaeGymaeJaeyypa0JaeGym % aeJaeGymaedabaGaeGinaqJaeyOpa4JaeGOmaiJaeyO0H4TaemOzay % 2aaeWaaeaacqaI0aanaiaawIcacaGLPaaacqGH9aqpqaaaaaaaaaWd % bmaalaaapaqaa8qadaGcaaWdaeaapeGaeGinaqJaey4kaSIaeGOmai % daleqaaOGaeyOeI0IaeG4mamdapaqaa8qacqaI0aancqGHRaWkcqaI % XaqmaaGaeyypa0ZaaSaaaeaadaGcaaqaaiabiAda2aWcbeaakiabgk % HiTiabiodaZaqaaiabiwda1aaaaeaacqGHshI3cqWGMbGzdaqadaqa % aiabgkHiTiabikdaYaGaayjkaiaawMcaaiabgUcaRiabiwda1iabdA % gaMnaabmaabaGaeGinaqdacaGLOaGaayzkaaGaeyypa0JaeGymaeJa % eGymaeJaey4kaSIaeGynauZaaSaaaeaadaGcaaqaaiabiAda2aWcbe % aakiabgkHiTiabiodaZaqaaiabiwda1aaacqGH9aqpcqaIXaqmcqaI % XaqmcqGHRaWkdaGcaaqaaiabiAda2aWcbeaakiabgkHiTiabiodaZi % abg2da9iabiIda4iabgUcaRmaakaaabaGaeGOnaydaleqaaaaaaa!8D38! \begin{array}{l} - 2 < 2 \Rightarrow f\left( { - 2} \right) = {\left( { - 2} \right)^2} - 3\left( { - 2} \right) + 1 = 11\\ 4 > 2 \Rightarrow f\left( 4 \right) = \frac{{\sqrt {4 + 2} - 3}}{{4 + 1}} = \frac{{\sqrt 6 - 3}}{5}\\ \Rightarrow f\left( { - 2} \right) + 5f\left( 4 \right) = 11 + 5\frac{{\sqrt 6 - 3}}{5} = 11 + \sqrt 6 - 3 = 8 + \sqrt 6 \end{array}\)