Tập nghiệm của bất phương trình: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac+ % gacaGGNbWaaSbaaSqaamaalaaabaGaaGymaaqaaiaaiodaaaaabeaa % kmaabmaabaGaamiEaiabgkHiTiaaiodaaiaawIcacaGLPaaacqGHsi % slcaaIXaGaeyOpa4JaaGimaaaa!421C! {\log _{\frac{1}{3}}}\left( {x - 3} \right) - 1 > 0\) có dạng (a;b). Khi đó giá trị a + 3b bằng ?
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Lời giải:
Báo saiĐiều kiện: x > 3
Bất phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaci % iBaiaac+gacaGGNbWaaSbaaSqaamaalaaabaGaaGymaaqaaiaaioda % aaaabeaakmaabmaabaGaamiEaiabgkHiTiaaiodaaiaawIcacaGLPa % aacqGH+aGpcaaIXaGaeyi1HSTaamiEaiabgkHiTiaaiodacqGH8aap % daWcaaqaaiaaigdaaeaacaaIZaaaaiabgsDiBlaadIhacqGH8aapda % WcaaqaaiaaigdacaaIWaaabaGaaG4maaaaaaa!50FF! \Leftrightarrow {\log _{\frac{1}{3}}}\left( {x - 3} \right) > 1 \Leftrightarrow x - 3 < \frac{1}{3} \Leftrightarrow x < \frac{{10}}{3}\)
So điều kiện \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiabg2 % da9maabmaabaGaaG4maiaacUdadaWcaaqaaiaaigdacaaIWaaabaGa % aG4maaaaaiaawIcacaGLPaaaaaa!3D18! S = \left( {3;\frac{{10}}{3}} \right)\), nên a + 3b = 13.