Có bao nhiêu điểm M thuộc đồ thị hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaamiEaiabgUcaRiaaikdaaeaacaWG4bGaeyOeI0Ia % aGymaaaaaaa!3D48! y = \frac{{x + 2}}{{x - 1}}\) sao cho khoảng cách từ M đến trục tung bằng hai lần khoảng cách từ M đến trục hoành.
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Lời giải:
Báo sai\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamytaiabgI % GiopaabmaabaGaam4qaaGaayjkaiaawMcaaiabgkDiElaad2eadaqa % daqaaiaad2gacaGG7aWaaSaaaeaacaWGTbGaey4kaSIaaGOmaaqaai % aad2gacqGHsislcaaIXaaaaaGaayjkaiaawMcaamaabmaabaGaamyB % aiabgcMi5kaaigdaaiaawIcacaGLPaaaaaa!4B3B! M \in \left( C \right) \Rightarrow M\left( {m;\frac{{m + 2}}{{m - 1}}} \right)\left( {m \ne 1} \right)\)
Theo bài ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizamaabm % aabaGaamytaiaacYcacaWGpbGaamyEaaGaayjkaiaawMcaaiabg2da % 9iaaikdacaWGKbWaaeWaaeaacaWGnbGaaiilaiaad+eacaWG4baaca % GLOaGaayzkaaGaeyi1HS9aaqWaaeaacaWGTbaacaGLhWUaayjcSdGa % eyypa0JaaGOmamaaemaabaWaaSaaaeaacaWGTbGaey4kaSIaaGOmaa % qaaiaad2gacqGHsislcaaIXaaaaaGaay5bSlaawIa7aaaa!53CF! d\left( {M,Oy} \right) = 2d\left( {M,Ox} \right) \Leftrightarrow \left| m \right| = 2\left| {\frac{{m + 2}}{{m - 1}}} \right|\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aaq % WaaeaacaWGTbWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaamyBaaGa % ay5bSlaawIa7aiabg2da9iaaikdadaabdaqaaiaad2gacqGHRaWkca % aIYaaacaGLhWUaayjcSdGaeyi1HS9aamqaaqaabeqaaiaad2gadaah % aaWcbeqaaiaaikdaaaGccqGHsislcaWGTbGaeyypa0JaaGOmaiaad2 % gacqGHRaWkcaaI0aaabaGaamyBamaaCaaaleqabaGaaGOmaaaakiab % gkHiTiaad2gacqGH9aqpcqGHsislcaaIYaGaamyBaiabgkHiTiaais % dacaGGOaGaamOvaiaad6eacaGGPaaaaiaawUfaaiabgsDiBpaadeaa % eaqabeaacaWGTbGaeyypa0JaaGinaaqaaiaad2gacqGH9aqpcqGHsi % slcaaIXaaaaiaawUfaaaaa!67EB! \Leftrightarrow \left| {{m^2} - m} \right| = 2\left| {m + 2} \right| \Leftrightarrow \left[ \begin{array}{l} {m^2} - m = 2m + 4\\ {m^2} - m = - 2m - 4(VN) \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} m = 4\\ m = - 1 \end{array} \right.\)
Vậy có 2 điểm M .