Tìm số phức z thỏa mãn |z - 2| = |z| và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WG6bGaey4kaSIaaGymaaGaayjkaiaawMcaamaabmaabaGabmOEayaa % raGaeyOeI0IaamyAaaGaayjkaiaawMcaaaaa!3E94! \left( {z + 1} \right)\left( {\bar z - i} \right)\) là số thực.
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Lời giải:
Báo saiGọi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOEaiabg2 % da9iaadIhacqGHRaWkcaWGPbGaamyEaaaa!3BC4! z = x + iy\) với \( x, y \in R\) ta có hệ phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaamaaemaabaGaamOEaiabgkHiTiaaikdaaiaawEa7caGLiWoacqGH % 9aqpdaabdaqaaiaadQhaaiaawEa7caGLiWoaaeaadaqadaqaaiaadQ % hacqGHRaWkcaaIXaaacaGLOaGaayzkaaWaaeWaaeaaceWG6bGbaeba % cqGHsislcaWGPbaacaGLOaGaayzkaaGaeyicI4SaeSyhHekaaiaawU % haaaaa!4D9A! \left\{ \begin{array}{l} \left| {z - 2} \right| = \left| z \right|\\ \left( {z + 1} \right)\left( {\bar z - i} \right) \in R \end{array} \right.\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aai % qaaqaabeqaamaabmaabaGaamiEaiabgkHiTiaaikdaaiaawIcacaGL % PaaadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaWG5bWaaWbaaSqabe % aacaaIYaaaaOGaeyypa0JaamiEamaaCaaaleqabaGaaGOmaaaakiab % gUcaRiaadMhadaahaaWcbeqaaiaaikdaaaaakeaadaqadaqaaiaadI % hacqGHRaWkcaaIXaGaey4kaSIaamyAaiaadMhaaiaawIcacaGLPaaa % daqadaqaaiaadIhacqGHsislcaWGPbGaamyEaiabgkHiTiaadMgaai % aawIcacaGLPaaacqGHiiIZcqWIDesOaaGaay5Eaaaaaa!584E! \Leftrightarrow \left\{ \begin{array}{l} {\left( {x - 2} \right)^2} + {y^2} = {x^2} + {y^2}\\ \left( {x + 1 + iy} \right)\left( {x - iy - i} \right) \in R \end{array} \right.\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aai % qaaqaabeqaaiaadIhacqGH9aqpcaaIXaaabaWaaeWaaeaacqGHsisl % caWG4bGaeyOeI0IaaGymaaGaayjkaiaawMcaamaabmaabaGaamyEai % abgUcaRiaaigdaaiaawIcacaGLPaaacqGHRaWkcaWG4bGaamyEaiab % g2da9iaaicdaaaGaay5Eaaaaaa!4A0B! \Leftrightarrow \left\{ \begin{array}{l} x = 1\\ \left( { - x - 1} \right)\left( {y + 1} \right) + xy = 0 \end{array} \right.\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aai % qaaqaabeqaaiaadIhacqGH9aqpcaaIXaaabaGaamyEaiabg2da9iab % gkHiTiaaikdaaaGaay5Eaaaaaa!3FDC! \Leftrightarrow \left\{ \begin{array}{l} x = 1\\ y = - 2 \end{array} \right.\)\(\Rightarrow z = 1 - 2i\)
Đề thi thử tốt nghiệp THPT QG môn Toán năm 2020
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