Cho hai số phức \(z,w\) thỏa mãn \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaamaaemaabaGaamOEaiabgkHiTiaaiodacqGHsislcaaIYaGaamyA % aaGaay5bSlaawIa7aiabgsMiJkaaigdaaeaadaabdaqaaiaadEhacq % GHRaWkcaaIXaGaey4kaSIaaGOmaiaadMgaaiaawEa7caGLiWoacqGH % KjYOdaabdaqaaiaadEhacqGHsislcaaIYaGaeyOeI0IaamyAaaGaay % 5bSlaawIa7aaaacaGL7baaaaa!5385! \left\{ \begin{array}{l} \left| {z - 3 - 2i} \right| \le 1\\ \left| {w + 1 + 2i} \right| \le \left| {w - 2 - i} \right| \end{array} \right.\) . Tìm giá trị nhỏ nhất \(P_{min}\) của biểu thức P = |z - w|.
Hãy suy nghĩ và trả lời câu hỏi trước khi xem đáp án
Lời giải:
Báo saiGiả sử z = a + bi ; w = x + yi ; \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WGHbGaaiilaiaadkgacaGGSaGaamiEaiaacYcacaWG5bGaeyicI4Sa % eSyhHekacaGLOaGaayzkaaaaaa!4049! \left( {a,b,x,y \in R} \right)\) . Ta có
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca % WG6bGaeyOeI0IaaG4maiabgkHiTiaaikdacaWGPbaacaGLhWUaayjc % SdGaeyizImQaaGymaaaa!40C6! \left| {z - 3 - 2i} \right| \le 1\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aae % WaaeaacaWGHbGaeyOeI0IaaG4maaGaayjkaiaawMcaamaaCaaaleqa % baGaaGOmaaaakiabgUcaRmaabmaabaGaamOyaiabgkHiTiaaikdaai % aawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGHKjYOcaaIXaaa % aa!45BA! \Leftrightarrow {\left( {a - 3} \right)^2} + {\left( {b - 2} \right)^2} \le 1\)
. Suy ra tập hợp điểm M biểu diễn số phức z là hình tròn tâm I ( 3;2), bán kính R = 1.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca % WG3bGaey4kaSIaaGymaiabgUcaRiaaikdacaWGPbaacaGLhWUaayjc % SdGaeyizIm6aaqWaaeaacaWG3bGaeyOeI0IaaGOmaiabgkHiTiaadM % gaaiaawEa7caGLiWoaaaa!4792! \left| {w + 1 + 2i} \right| \le \left| {w - 2 - i} \right|\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aae % WaaeaacaWG4bGaey4kaSIaaGymaaGaayjkaiaawMcaamaaCaaaleqa % baGaaGOmaaaakiabgUcaRmaabmaabaGaamyEaiabgUcaRiaaikdaai % aawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGHKjYOdaqadaqa % aiaadIhacqGHsislcaaIYaaacaGLOaGaayzkaaWaaWbaaSqabeaaca % aIYaaaaOGaey4kaSYaaeWaaeaacaWG5bGaeyOeI0IaaGymaaGaayjk % aiaawMcaamaaCaaaleqabaGaaGOmaaaakiabgsDiBlaadIhacqGHR % WkcaWG5bGaeyizImQaaGimaaaa!57E3! \Leftrightarrow {\left( {x + 1} \right)^2} + {\left( {y + 2} \right)^2} \le {\left( {x - 2} \right)^2} + {\left( {y - 1} \right)^2} \Leftrightarrow x + y \le 0\)
Suy ra tập hợp điểm N biểu diễn số phức w là nửa mặt phẳng giới hạn bởi đường thẳng \(\Delta: x + y = 0\) không chứa I
Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizamaabm % aabaGaamysaiaacYcacqqHuoaraiaawIcacaGLPaaacqGH9aqpdaWc % aaqaaiaaiwdaaeaadaGcaaqaaiaaikdaaSqabaaaaaaa!3DF6! d\left( {I,\Delta } \right) = \frac{5}{{\sqrt 2 }}\). Gọi H là hình chiếu của I trên \(\Delta\) .
Khi đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca % WG6bGaeyOeI0Iaam4DaaGaay5bSlaawIa7aiabg2da9iaad2eacaWG % obGaeyyzImRaamizamaabmaabaGaamysaiaacYcacqqHuoaraiaawI % cacaGLPaaacqGHsislcaWGsbGaeyypa0ZaaSaaaeaacaaI1aWaaOaa % aeaacaaIYaaaleqaaaGcbaGaaGOmaaaacqGHsislcaaIXaaaaa!4CA3! \left| {z - w} \right| = MN \ge d\left( {I,\Delta } \right) - R = \frac{{5\sqrt 2 }}{2} - 1\). Suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuamaaBa % aaleaaciGGTbGaaiyAaiaac6gaaeqaaOGaeyypa0ZaaSaaaeaacaaI % 1aWaaOaaaeaacaaIYaaaleqaaaGcbaGaaGOmaaaacqGHsislcaaIXa % aaaa!3EEB! {P_{\min }} = \frac{{5\sqrt 2 }}{2} - 1\).
Đề thi thử tốt nghiệp THPT QG môn Toán năm 2020
Tuyển chọn số 2