Cho hai số thực dương a,b thỏa mãn \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCsPYR2xbba9asFD0dXdHaVhbbf9v8qqaqFr0xc9pk % 0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9 % Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaacaWGHbWaaW % baaSqabeaacaaIYaaaaOGaey4kaSIaamOyamaaCaaaleqabaGaaGOm % aaaakiabg6da+iaaigdaaaa!3EE1! {a^2} + {b^2} > 1\) và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCsPYR2xbba9asFD0dXdHaVhbbf9v8qqaqFr0xc9pk % 0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9 % Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaciGGSbGaai % 4BaiaacEgadaWgaaWcbaGaamyyamaaCaaameqabaGaaGOmaaaaliab % gUcaRiaadkgadaahaaadbeqaaiaaikdaaaaaleqaaOWaaeWaaeaaca % WGHbGaey4kaSIaamOyaaGaayjkaiaawMcaaiabgwMiZkaaigdaaaa!46E1! {\log _{{a^2} + {b^2}}}\left( {a + b} \right) \ge 1\). Tìm giá trị lớn nhất của biểu thức P = 2a + 4b - 3.

A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCsPYR2xbba9asFD0dXdHaVhbbf9v8qqaqFr0xc9pk % 0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9 % Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaadaWcaaqaam % aakaaabaGaaGymaiaaicdaaSqabaaakeaacaaIYaaaaaaa!3AF0! \frac{{\sqrt {10} }}{2}\)

B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCsPYR2xbba9asFD0dXdHaVhbbf9v8qqaqFr0xc9pk % 0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9 % Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaadaGcaaqaai % aaigdacaaIWaaaleqaaaaa!3A1A! \sqrt {10} \)

C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCsPYR2xbba9asFD0dXdHaVhbbf9v8qqaqFr0xc9pk % 0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9 % Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaadaWcaaqaai % aaikdadaGcaaqaaiaaigdacaaIWaaaleqaaaGcbaGaaGOmaaaaaaa!3BAC! \frac{{2\sqrt {10} }}{2}\)

D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCsPYR2xbba9asFD0dXdHaVhbbf9v8qqaqFr0xc9pk % 0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9 % Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaadaWcaaqaai % aaigdaaeaadaGcaaqaaiaaigdacaaIWaaaleqaaaaaaaa!3AE5! \frac{1}{{\sqrt {10} }}\)

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Môn: Toán Lớp 12

Chủ đề: Hàm Số Lũy Thừa Hàm Số Mũ Và Hàm Số Lôgarit

Bài: Hàm số lũy thừa

Lời giải:

Báo sai

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCsPYR2xbba9asFD0dXdHaVhbbf9v8qqaqFr0xc9pk % 0xbba9q8WqFfea0-yr0RYxir-Jbba9q8aq0-yq-He9q8qqQ8frFve9 % Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaciGGSbGaai % 4BaiaacEgadaWgaaWcbaGaamyyamaaCaaameqabaGaaGOmaaaaliab % gUcaRiaadkgadaahaaadbeqaaiaaikdaaaaaleqaaOWaaeWaaeaaca % WGHbGaey4kaSIaamOyaaGaayjkaiaawMcaaiabgwMiZkaaigdacqGH % uhY2caWGHbWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaamOyamaaCa % aaleqabaGaaGOmaaaakiabgsMiJkaadggacqGHRaWkcaWGIbGaeyi1 % HS9aaeWaaeaacaWGHbGaeyOeI0YaaSaaaeaacaaIXaaabaGaaGOmaa % aaaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGHRaWkdaqa % daqaaiaadkgacqGHsisldaWcaaqaaiaaigdaaeaacaaIYaaaaaGaay % jkaiaawMcaamaaCaaaleqabaGaaGOmaaaakiabgsMiJoaalaaabaGa % aGymaaqaaiaaikdaaaaaaa!645D! {\log _{{a^2} + {b^2}}}\left( {a + b} \right) \ge 1 \Leftrightarrow {a^2} + {b^2} \le a + b \Leftrightarrow {\left( {a - \frac{1}{2}} \right)^2} + {\left( {b - \frac{1}{2}} \right)^2} \le \frac{1}{2}\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuaiabg2 % da9iaaikdadaqadaqaaiaadggacqGHsisldaWcaaqaaiaaigdaaeaa % caaIYaaaaaGaayjkaiaawMcaaiabgUcaRiaaisdadaqadaqaaiaadk % gacqGHsisldaWcaaqaaiaaigdaaeaacaaIYaaaaaGaayjkaiaawMca % aiabgsMiJoaakaaabaWaaeWaaeaacaaIYaWaaWbaaSqabeaacaaIYa % aaaOGaey4kaSIaaGinamaaCaaaleqabaGaaGOmaaaaaOGaayjkaiaa % wMcaamaadmaabaWaaeWaaeaacaWGHbGaeyOeI0YaaSaaaeaacaaIXa % aabaGaaGOmaaaaaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGc % cqGHRaWkdaqadaqaaiaadkgacqGHsisldaWcaaqaaiaaigdaaeaaca % aIYaaaaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaaaOGaay5w % aiaaw2faaaWcbeaaaaa!5A0E! P= 2\left( {a - \frac{1}{2}} \right) + 4\left( {b - \frac{1}{2}} \right) \le \sqrt {\left( {{2^2} + {4^2}} \right)\left[ {{{\left( {a - \frac{1}{2}} \right)}^2} + {{\left( {b - \frac{1}{2}} \right)}^2}} \right]} \)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyizIm6aaO % aaaeaacaaIYaGaaGimaiaac6cadaWcaaqaaiaaigdaaeaacaaIYaaa % aaWcbeaakiabg2da9maakaaabaGaaGymaiaaicdaaSqabaaaaa!3E13! \le \sqrt {20.\frac{1}{2}} = \sqrt {10} \)