{"id":1129,"date":"2025-10-16T14:47:52","date_gmt":"2025-10-16T14:47:52","guid":{"rendered":"https:\/\/blog.ngocha.biz\/?p=1129"},"modified":"2026-05-27T18:22:43","modified_gmt":"2026-05-27T18:22:43","slug":"song-elliott","status":"publish","type":"post","link":"https:\/\/blog.ngocha.biz\/?p=1129","title":{"rendered":"S\u00f3ng Elliott"},"content":{"rendered":"

Tr\u01b0\u1edbc h\u1ebft h\u00e3y nh\u00ecn \u0111\u1ed3 th\u1ecb ng\u00e0y c\u1ee7a HPG:<\/p>\n

\"\"<\/figure>\n

Nh\u00ecn \u0111\u1ed3 th\u1ecb n\u00e0y b\u1ea1n c\u00f3 nh\u00ecn th\u1ea5y c\u00f3 quy lu\u1eadt g\u00ec kh\u00f4ng? T\u1ea1i sao n\u00f3 l\u1ea1i t\u1ea1o c\u00e1c \u0111\u1ec9nh 1,3,5; xen k\u1ebd v\u1edbi \u0111\u00f3 l\u00e0 c\u00e1c l\u1ea7n \u0111i\u1ec1u ch\u1ec9nh t\u1ea1o \u0111\u00e1y 2,4 ? N\u1ebfu th\u1ef1c s\u1ef1 c\u00f3 quy lu\u1eadt th\u00ec b\u1ea1n c\u00f3 th\u1ec3 bi\u1ebft tr\u01b0\u1edbc \u0111\u1ec3 t\u1eeb \u0111\u00f3 t\u1ed1i \u01b0u h\u00f3a vi\u1ec7c ra quy\u1ebft \u0111\u1ecbnh.<\/p>\n

Quy lu\u1eadt \u0111\u00f3 l\u00e0 s\u00f3ng Ellliott<\/strong><\/p>\n

S\u01a1 \u0111\u1ed3:<\/strong><\/p>\n

\"\"<\/figure>\n

M\u1ed9t chu k\u1ef3 ho\u00e0n ch\u1ec9nh s\u00f3ng Elliott bao g\u1ed3m s\u00f3ng \u0111\u1ea9y l\u00ean v\u00e0 s\u00f3ng \u0111i\u1ec1u ch\u1ec9nh. S\u00f3ng \u0111\u1ea9y bao g\u1ed3m ba s\u00f3ng t\u0103ng 1,3,5; hai s\u00f3ng \u0111i\u1ec1u ch\u1ec9nh 2,4. S\u00f3ng \u0111\u1ea9y t\u1ea1o ra ba \u0111\u1ec9nh s\u00f3ng. S\u00f3ng \u0111i\u1ec1u ch\u1ec9nh l\u00e0 s\u00f3ng A,B,C.<\/p>\n

Chu k\u1ef3 s\u00f3ng Elliott n\u00e0y c\u00f3 th\u1ec3 n\u1eb1m trong m\u1ed9t s\u00f3ng l\u1edbn h\u01a1n trong \u0111\u00f3 c\u1ea3 chu k\u1ef3 ch\u1ec9 l\u00e0 s\u00f3ng 1. Ho\u1eb7c m\u1ed7i s\u00f3ng trong chu k\u1ef3 n\u00e0y n\u00f3 c\u0169ng ch\u1ee9a m\u1ed9t chu k\u1ef3. T\u1ea1i \u0111i\u1ec3m C kh\u00f4ng c\u00f3 ngh\u0129a l\u00e0 n\u00f3 c\u1eafm \u0111\u1ea7u xu\u1ed1ng m\u00e3i m\u00e0 C \u00e1m ch\u1ec9 r\u1eb1ng cp \u0111\u00e3 k\u1ebft th\u00fac m\u1ed9t chu k\u1ef3. Nh\u1eefng g\u00ec ti\u1ebfp sau chu k\u1ef3 n\u00e0y c\u00f2n ph\u1ee5 thu\u1ed9c v\u00e0o nhi\u1ec1u y\u1ebfu t\u1ed1 kh\u00e1c.<\/p>\n

Trong m\u1ed9t xu h\u01b0\u1edbng gi\u1ea3m n\u00f3 c\u0169ng c\u00f3 th\u1ec3 gi\u1ea3m theo s\u00f3ng Eliot l\u1ed9n ng\u01b0\u1ee3c. V\u00ec v\u1eady ta c\u00f3 th\u1ec3 \u00e1p d\u1ee5ng cho c\u1ea3 m\u1ed9t xu h\u01b0\u1edbng t\u0103ng l\u1eabn xu h\u01b0\u1edbng gi\u1ea3m.<\/p>\n

Ba Quy t\u1eafc:<\/strong><\/mark><\/p>\n

Quy t\u1eafc 1:<\/strong> S\u00f3ng 2 khi gi\u1ea3m kh\u00f4ng \u0111\u01b0\u1ee3c th\u1ea5p h\u01a1n ch\u00e2n s\u00f3ng 1. N\u1ebfu s\u00f3ng 2 xu\u1ed1ng th\u1ea5p h\u01a1n ch\u00e2n s\u00f3ng 1 th\u00ec ta \u0111\u00e3 x\u00e1c \u0111\u1ecbnh s\u00f3ng sai. N\u00f3 th\u01b0\u1eddng d\u1eebng l\u1ea1i trong v\u00f9ng Fibonacci 0.5 v\u00e0 0,618 c\u1ee7a s\u00f3ng 1; n\u1ebfu xu h\u01b0\u1edbng m\u1ea1nh n\u00f3 ch\u1ec9 h\u1ed3i l\u1ea1i v\u00f9ng 0,382<\/p>\n

\"\"<\/figure>\n

Quy t\u1eafc 2: <\/strong>S\u00f3ng 3 kh\u00f4ng bao gi\u1edd l\u00e0 s\u00f3ng ng\u1eafn nh\u1ea5t trong 3 s\u00f3ng 1,3,5. S\u00f3ng 3 l\u00e0 s\u00f3ng c\u00f3 xu th\u1ebf m\u1ea1nh nh\u1ea5t trong 3 s\u00f3ng 1,3,5 v\u00e0 th\u01b0\u1eddng c\u00f3 \u0111\u1ed9 d\u00e0i b\u1eb1ng s\u00f3ng 1 ho\u1eb7c s\u00f3ng 5.<\/p>\n

Quy t\u1eafc 3:<\/strong> S\u00f3ng 4 kh\u00f4ng bao gi\u1edd \u0111\u01b0\u1ee3c vi ph\u1ea1m v\u00e0o v\u00f9ng s\u00f3ng 1 (kh\u00f4ng th\u1ea5p h\u01a1n \u0111\u1ec9nh s\u00f3ng 1)<\/p>\n

1.Pha \u0111\u1ea9y:<\/mark><\/strong><\/p>\n

Nh\u1eadn bi\u1ebft s\u00f3ng 1<\/strong><\/p>\n

Ai c\u0169ng mu\u1ed1n mua \u1edf ch\u00e2n s\u00f3ng 1 sau \u0111\u00f3 gi\u1eef cho t\u1edbi s\u00f3ng 5 \u0111\u1ec3 t\u1ed1i \u01b0u h\u00f3a l\u1ee3i nhu\u1eadn. Ch\u00fang ta c\u00f3 hai v\u1ea5n \u0111\u1ec1 \u1edf \u0111\u00e2y: L\u00e0m sao bi\u1ebft \u0111\u01b0\u1ee3c l\u00e0 chu\u1ea9n b\u1ecb hay b\u1eaft \u0111\u1ea7u v\u00e0o s\u00f3ng 1? v\u00e0 l\u00e0m sao bi\u1ebft \u0111\u01b0\u1ee3c n\u00f3 s\u1ebd c\u00f3 3 l\u1ea7n s\u00f3ng t\u0103ng \u0111\u1ec3 m\u00e0 gi\u1eef t\u1edbi s\u00f3ng 5? <\/p>\n

Tr\u01b0\u1edbc khi b\u1eaft \u0111\u1ea7u s\u00f3ng 1 th\u00ec cp c\u00f3 th\u1ec3 \u0111ang trong m\u1ed9t xu h\u01b0\u1edbng gi\u1ea3m r\u00f5 r\u00e0ng; khi n\u00f3 t\u0103ng ng\u01b0\u1eddi ta s\u1ebd nghi ng\u1edd r\u1eb1ng \u0111\u00f3 l\u00e0 m\u1ed9t c\u00fa h\u1ed3i t\u1ea1m th\u1eddi \u0111\u1ec3 r\u1ed3i l\u1ea1i ti\u1ebfp t\u1ee5c xu h\u01b0\u1edbng gi\u1ea3m. Ho\u1eb7c tr\u01b0\u1edbc khi c\u00f3 s\u00f3ng 1 th\u00ec cp \u0111i ngang c\u00f3 khi c\u1ea3 n\u0103m tr\u1eddi; khi cp t\u0103ng ng\u01b0\u1eddi ta s\u1ebd nghi ng\u1edd r\u1eb1ng n\u00f3 t\u0103ng ng\u1eabu nhi\u00ean do m\u1ed9t th\u1eb1ng h\u00e2m n\u00e0o \u0111\u00f3 nh\u1ea9y v\u00e0o mua nhi\u1ec1u; r\u1ed3i sau \u0111\u00f3 n\u00f3 l\u1ea1i ti\u1ebfp t\u1ee5c sideway; mua gi\u1edd l\u00e0 b\u1ecb g\u0103m v\u1ed1n.<\/p>\n

V\u00ec v\u1eady c\u00e1ch l\u00e0m c\u1ee7a \u0111a s\u1ed1 n\u0111t l\u00e0 sau khi s\u00f3ng 1 ho\u00e0n th\u00e0nh, ng\u01b0\u1eddi ta m\u1edbi nh\u1eadn ra \u0111\u00f3 l\u00e0 s\u00f3ng 1, h\u1ecd \u0111\u1ee3i c\u00fa \u0111i\u1ec1u ch\u1ec9nh c\u1ee7a s\u00f3ng 2 \u0111\u1ec3 nh\u1ea9y v\u00e0o nh\u1eb1m \u0103n s\u00f3ng 3 v\u00e0 5.<\/p>\n

S\u00f3ng 1 h\u00ecnh th\u00e0nh kh\u00f4ng ph\u1ea3i do nh\u1ecf l\u1ebb v\u00ec nh\u1ecf l\u1ebb kh\u00f4ng \u0111\u1ee7 ti\u1ec1m l\u1ef1c l\u00e0m th\u1ebf<\/strong>. C\u00e1c tay to s\u1ebd gom h\u00e0ng tr\u01b0\u1edbc khi b\u1eaft \u0111\u1ea7u ch\u00e2n s\u00f3ng 1 (giai \u0111o\u1ea1n t\u00edch l\u0169y ho\u1eb7c gi\u1ea3m t\u1edbi ch\u00e1n n\u1ea3n) sau \u0111\u00f3 h\u1ecd nh\u1eb9 nh\u00e0ng \u0111\u1ea9y l\u00ean; trong m\u1ed9t tr\u1ea1ng th\u00e1i ch\u00e1n n\u1ea3n cp s\u1ebd kh\u00f4ng b\u1ecb \u0111\u1ec3 \u00fd v\u00e0 ng\u01b0\u1eddi c\u00f2n cp c\u00f3 khi c\u0169ng ch\u1eb3ng quan t\u00e2m v\u00ec v\u1eady v\u1edbi m\u1ed9t kh\u1ed1i l\u01b0\u1ee3ng v\u1eeba ph\u1ea3i h\u1ecd \u0111\u00e3 c\u00f3 th\u1ec3 kh\u1edfi t\u1ea1o con s\u00f3ng 1. Kh\u1ed1i l\u01b0\u1ee3ng n\u00e0y s\u1ebd ph\u1ea3i cao \u0111\u1ed9t bi\u1ebfn so v\u1edbi trung b\u00ecnh 20 phi\u00ean.<\/p>\n

S\u00f3ng 2 l\u00e0 khi c\u00e1c n\u0111t nh\u1ecf l\u1ebb c\u1ea7m h\u00e0ng ch\u1ed1t l\u00e3i trong khi tay to l\u1ea1i mua v\u00e0o. Ch\u1ec9 c\u00f3 r\u1ea5t \u00edt n\u0111t nhi\u1ec1u kinh nghi\u1ec7m nh\u1eadn ra \u0111\u1ec3 tranh th\u1ee7 mua h\u00e0ng \u1edf s\u00f3ng 2. Sau \u0111\u00f3 tay to l\u1ea1i \u0111\u1ea9y gi\u00e1 t\u0103ng ti\u1ebfp t\u1ea1o th\u00e0nh s\u00f3ng 3. S\u00f3ng 3 l\u00e0 l\u00fac m\u00e0 h\u1ea7u h\u1ebft nh\u1eadn ra \u0111\u00f3 l\u00e0 …. s\u00f3ng 3 v\u00ec v\u1eady l\u01b0\u1ee3ng cung \u00edt trong khi l\u01b0\u1ee3ng c\u1ea7u nhi\u1ec1u. CP t\u0103ng nhanh d\u1ed1c \u0111\u1ee9ng. V\u00ec v\u1eady s\u00f3ng 3 c\u00f3 \u0111\u1ed9ng l\u1ef1c m\u1ea1nh nh\u1ea5t.<\/p>\n

S\u00f3ng 4 l\u00e0 l\u00fac m\u00e0 tay to b\u1eaft \u0111\u1ea7u b\u00e1n ra; th\u1ec9nh tho\u1ea3ng s\u1ebd c\u00fa t\u0103ng m\u1ea1nh \u0111\u1ec3 t\u1ea1o t\u00e2m l\u00fd FOMO nh\u01b0ng sau \u0111\u00f3 gi\u00e1 l\u1ea1i gi\u1ea3m v\u1ec1. Trong s\u00f3ng 2 nh\u1ecf l\u1ebb kh\u00f4ng d\u00e1m mua v\u00ec v\u1eady h\u1ecd kh\u00f4ng \u0103n \u0111\u01b0\u1ee3c s\u00f3ng 3 sinh ra tr\u1ea1ng th\u00e1i ti\u1ebfc r\u1ebb n\u00ean khi s\u00f3ng 4 x\u1ea3y b\u1eaft \u0111\u1ea7u suy ngh\u0129 c\u00f3 n\u00ean mua kh\u00f4ng. Nh\u1eefng c\u00fa t\u0103ng m\u1ea1nh sau \u0111\u00f3 (r\u1ed3i l\u1ea1i gi\u1ea3m v\u1ec1) k\u00edch th\u00edch h\u1ecd h\u00e0nh \u0111\u1ed9ng. Cu\u1ed1i c\u00f9ng kh\u00f4ng ch\u1ecbu \u0111\u01b0\u1ee3c h\u1ecd s\u1ebd mua. Gi\u00e1 ti\u1ebfp t\u1ee5c loanh quanh cho t\u1edbi khi tay to tho\u00e1t h\u1ebft h\u00e0ng. Nh\u01b0 v\u1eady tay to kh\u00f4ng tho\u00e1t h\u00e0ng \u1edf s\u00f3ng 5 m\u00e0 h\u1ecd \u0111\u00e3 tho\u00e1t h\u1ebft \u1edf s\u00f3ng 4.<\/p>\n

S\u00f3ng 5: \u0111\u01b0\u1ee3c t\u1ea1o b\u1edfi s\u1ef1 Fomo c\u1ee7a nh\u1ecf l\u1ebb l\u00e0 ch\u00ednh n\u00ean s\u1ebd t\u1ea1o s\u1ef1 ph\u00e2n k\u1ef3 trong c\u00e1c ch\u1ec9 b\u00e1o. Gi\u00e1 t\u0103ng nh\u01b0ng k\u00e8m kh\u1ed1i l\u01b0\u1ee3ng gi\u1ea3m d\u1ea7n; gi\u00e1 t\u0103ng nh\u01b0ng RSI gi\u1ea3m; \u0111\u01b0\u1eddng MACD gi\u1ea3m. T\u1ea5t nhi\u00ean c\u0169ng s\u1ebd c\u00f3 tay to c\u1ea7m h\u00e0ng cho t\u1edbi h\u1ebft s\u00f3ng 5 v\u00e0 ra h\u00e0ng t\u1ea1i \u0111\u1ec9nh r\u1ed3i ra h\u1ebft trong pha \u0111i\u1ec1u ch\u1ec9nh. ABC.<\/p>\n

S\u1ef1 m\u1edf r\u1ed9ng c\u1ee7a s\u00f3ng<\/strong><\/p>\n

S\u00f3ng 1,3,5 l\u00e0 s\u00f3ng t\u0103ng nh\u01b0ng trong m\u1ed9t s\u00f3ng t\u0103ng c\u0169ng c\u00f3 th\u1ec3 c\u00f3 c\u00e1c pha \u0111i\u1ec1u ch\u1ec9nh kh\u00f4ng ph\u1ea3i n\u00f3 t\u0103ng m\u1ed9t m\u1ea1ch theo ki\u1ec3u ng\u00e0y n\u00e0o c\u0169ng xanh. Trong 3 s\u00f3ng n\u00e0y s\u1ebd c\u00f3 m\u1ed9t s\u00f3ng bao g\u1ed3m nhi\u1ec1u s\u00f3ng nh\u1ecf \u1edf trong.<\/p>\n

N\u1ebfu chi\u1ec1u cao s\u00f3ng 3 ng\u1eafn h\u01a1n nhi\u1ec1u so v\u1edbi chi\u1ec1u cao s\u00f3ng 1 th\u00ec kh\u1ea3 n\u0103ng ta \u0111ang \u1edf trong m\u1ed9t pha \u0111i\u1ec1u ch\u1ec9nh nh\u00f3 c\u1ee7a s\u00f3ng 3 m\u00e0 kh\u00f4ng ph\u1ea3i \u0111ang \u1edf \u0111\u1ec9nh c\u1ee7a s\u00f3ng 3 v\u00ec s\u00f3ng 3 l\u00e0 s\u00f3ng m\u1ea1nh nh\u1ea5t n\u00ean n\u00f3 ph\u1ea3i cao h\u01a1n chi\u1ec1u cao c\u1ee7a s\u00f3ng 1. S\u00f3ng 3 th\u01b0\u1eddng c\u00f3 chi\u1ec1u cao b\u1eb1ng 1,618 chi\u1ec1u cao c\u1ee7a s\u00f3ng 1 (Fibonacci). S\u00f3ng 3 s\u1ebd c\u00f3 chi\u1ec1u cao theo c\u00e1c m\u1ee9c c\u1ee7a s\u00f3ng 1 l\u00e0 1,618; 200%; 261,8%; 300%; 361,8%; 400%.<\/p>\n

Khi s\u00f3ng 1 h\u00ecnh th\u00e0nh ta s\u1ebd nh\u00ecn xem n\u00f3 c\u00f3 ph\u1ea3i s\u00f3ng b\u00ecnh th\u01b0\u1eddng hay l\u00e0 n\u00f3 bao g\u1ed3m m\u1ed9t v\u00e0i s\u00f3ng nh\u1ecf \u1edf trong. T\u1ea1i s\u00f3ng 3 ta c\u0169ng xem n\u00f3 c\u00f3 bao g\u1ed3m nhi\u1ec1u s\u00f3ng nh\u1ecf kh\u00f4ng; tr\u00e1nh b\u00e1n ra m\u1ea5t h\u00e0ng trong m\u1ed9t s\u00f3ng \u0111i\u1ec1u ch\u1ec9nh gi\u1ea3. N\u1ebfu s\u00f3ng 1 kh\u00f4ng c\u00f3 s\u00f3ng nh\u1ecf \u0111i\u1ec1u ch\u1ec9nh, s\u00f3ng 3 b\u1eb1ng Fibonacci 1,618 c\u1ee7a s\u00f3ng 1 th\u00ec kh\u1ea3 n\u0103ng r\u1ea5t cao s\u00f3ng 5 s\u1ebd c\u00f3 s\u00f3ng nh\u1ecf ph\u00eda trong (3 ho\u1eb7c 5)<\/p>\n

2. Pha \u0111i\u1ec1u ch\u1ec9nh<\/mark><\/strong><\/p>\n

S\u00f3ng \u0111i\u1ec1u ch\u1ec9nh h\u00ecnh Z<\/strong><\/p>\n

S\u00f3ng \u0111i\u1ec1u ch\u1ec9nh l\u00e0 ABC c\u00f3 th\u1ec3 l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng nh\u01b0 m\u1eabu v\u00e0 c\u0169ng c\u00f3 th\u1ec3 c\u00f3 c\u00e1c h\u00ecnh d\u1ea1ng kh\u00e1c. Ch\u00fang ta bi\u1ebft r\u1eafng trong s\u00f3ng \u0111\u1ea9y l\u00ean \u1edf xu h\u01b0\u1edbng \u0111i l\u00ean th\u00ec m\u1ecdi th\u1ee9 c\u00f3 k\u1ecbch b\u1ea3n v\u00e0 k\u1ecbch b\u1ea3n n\u00e0y \u0111\u01b0\u1ee3c \u0111i\u1ec1u h\u00e0nh b\u1edfi tay to n\u00ean m\u1ecdi th\u1ee9 r\u1ea5t ng\u0103n l\u1eafp r\u00f5 r\u00e0ng. C\u00f2n trong s\u00f3ng \u0111i\u1ec1u ch\u1ec9nh ch\u1ec9 c\u00f2n l\u1ea1i m\u1ed9t \u00edt h\u00e0ng c\u1ee7a tay to (n\u1ebfu h\u1ecd ch\u01b0a tho\u00e1t h\u1ebft) v\u00e0 nh\u1ecf l\u1ebb v\u00ec v\u1eady s\u00f3ng \u0111i\u1ec1u ch\u1ec9nh s\u1ebd kh\u00f3 th\u1ec3 \u0111\u00fang c\u00f4ng th\u1ee9c.<\/p>\n

T\u1eeb \u0111\u1ec9nh 5 gi\u00e1 b\u1ecb \u0111\u1ea9y xu\u1ed1ng A r\u1ed3i h\u1ed3i l\u00ean B v\u00e0 gi\u1ea3m xu\u1ed1ng C. Trong m\u1ed9t s\u00f3ng c\u00f3 th\u1ec3 c\u00f3 nhi\u1ec1u s\u00f3ng nh\u1ecf k\u00e9o l\u00ean \u0111\u1ec3 t\u1ea1o t\u00e2m l\u00fd Fomo c\u1ee7a nh\u1eefng ng\u01b0\u1eddi c\u1ea7m ti\u1ec1n b\u00ean ngo\u00e0i; v\u00e0 s\u1ef1 ch\u1ea7n ch\u1edd kh\u00f4ng mu\u1ed1n b\u00e1n c\u1ee7a nh\u1eefng ng\u01b0\u1eddi c\u1ea9m cp. N\u00f3 c\u00f3 th\u1ec3 \u0111i zigzg h\u00ecnh ch\u1eef Z.<\/p>\n

\"\"<\/figure>\n

S\u00f3ng \u0111i\u1ec1u ch\u1ec9nh N<\/strong><\/p>\n

Tr\u01b0\u1eddng h\u1ee3p 1: L\u1ea7n n\u00e0y B thay v\u00ec h\u1ed3i th\u1ea5p h\u01a1n \u0111\u1ec9nh 5 th\u00ec n\u00f3 h\u1ed3i b\u1eb1ng \u0111\u1ec9nh 5 t\u1ea1o m\u00f4 h\u00ecnh hai \u0111\u1ec9nh<\/p>\n

Tr\u01b0\u1eddng h\u1ee3p 2: N\u00f3 h\u1ed3i cao h\u01a1n c\u1ea3 \u0111\u1ec9nh 5 t\u1ea1o s\u1ef1 nh\u1ea7m l\u1eabn r\u1eb1ng \u0111\u00e2y m\u1edbi l\u00e0 \u0111\u1ec9nh 5. V\u00ec nh\u1ea7m l\u1eabn n\u00e0y khi k\u00e9o Fibonacci t\u1eeb \u0111\u1ec9nh xu\u1ed1ng \u0111\u00e1y d\u1ec5 ch\u1ecdn nh\u1ea7m B l\u00e0 \u0111\u1ec9nh thay v\u00ec ch\u1ecdn 5 l\u00e0 \u0111\u1ec9nh. Trong v\u00ed d\u1ee5 \u1edf ph\u00eda tr\u00ean, HPG c\u00f3 v\u1ebb \u0111ang \u1edf trong tr\u01b0\u1eddng h\u1ee3p 2 khi B cao h\u01a1n \u0111\u1ec9nh 5.<\/p>\n

\"\"<\/figure>\n

T\u01b0\u01a1ng t\u1ef1 nh\u01b0 h\u00ecnh Z th\u00ec s\u00f3ng A,B,C c\u0169ng c\u00f3 th\u1ec3 bao g\u1ed3m trong \u0111\u00f3 nhi\u1ec1u con s\u00f3ng con. Trong s\u00f3ng A c\u00f3 th\u1ec3 c\u00f3 s\u00f3ng h\u1ed3i nh\u1eb9 r\u1ed3i l\u1ea1i \u0111i ti\u1ebfp xu\u1ed1ng A. Trong s\u00f3ng B c\u00f3 m\u1ed9t s\u00f3ng \u0111i\u1ec1u ch\u1ec9nh nh\u1ecf r\u1ed3i \u0111i ti\u1ebfp l\u00ean B. Nh\u1eefng s\u00f3ng h\u1ed3i v\u00e0 \u0111i\u1ec1u ch\u1ec9nh n\u00e0y m\u1ee5c \u0111\u00edch l\u00e0 \u0111\u1ec3 d\u1ee5 g\u00e0 nh\u1ea9y v\u00e0o; ch\u00fang ta c\u1ea7n t\u1ec9nh t\u00e1o quan s\u00e1t k\u1ebft h\u1ee3p d\u00f9ng c\u00f4ng c\u1ee5 Fibonacci. Nh\u1ea5n m\u1ea1nh l\u00e0 trong s\u00f3ng \u0111i\u1ec1u ch\u1ec9nh ABC b\u1ea1n n\u00ean ki\u00ean nh\u1eabn \u0111\u1ee9ng ngo\u00e0i nh\u00ecn tr\u1eeb khi l\u00e0 chuy\u00ean gia. <\/p>\n

S\u00f3ng \u0111i\u1ec1u ch\u1ec9nh d\u1ea1ng tam gi\u00e1c<\/strong><\/p>\n

S\u00f3ng \u0111i\u1ec1u ch\u1ec9nh tam gi\u00e1c \u0111i l\u00ean:<\/p>\n

\"\"<\/figure>\n

S\u00f3ng \u0111i\u1ec1u ch\u1ec9nh tam gi\u00e1c \u0111i xu\u1ed1ng:<\/p>\n

\"\"<\/figure>\n

S\u00f3ng abcd n\u00e0y b\u1ea3n th\u00e2n trong n\u00f3 c\u00f3 th\u1ec3 ch\u1ee9a c\u00e1c s\u00f3ng nh\u1ecf. Ta s\u1ebd d\u00f9ng c\u00e1c \u0111\u01b0\u1eddng d\u1eabn \u0111\u1ec3 nh\u1eadn di\u1ec7n. M\u1ed9t cp trong xu h\u01b0\u1edbng t\u0103ng kh\u00f4ng c\u00f3 ngh\u0129a n\u00f3 ho\u00e0n th\u00e0nh 5 s\u00f3ng \u0111\u1ea9y r\u1ed3i c\u1eafm \u0111\u1ea7u m\u00e0 c\u00f3 th\u1ec3 chuy\u1ec3n sang t\u00edch l\u0169y \u0111\u1ec3 ti\u1ebfp t\u1ee5c m\u1ed9t chu k\u1ef3 t\u0103ng m\u1edbi (T\u01b0\u01a1ng t\u1ef1 v\u1edbi m\u1ed9t xu h\u01b0\u1edbng gi\u1ea3m)<\/p>\n

Vi\u1ec7c c\u1ee7a ch\u00fang ta l\u00e0 tho\u00e1t ra trong s\u00f3ng 5 r\u1ed3i \u0111\u1ee3i ch\u1edd t\u00edn hi\u1ec7u x\u00e1c nh\u1eadn n\u00f3 break l\u00ean hay break xu\u1ed1ng \u0111\u1ec3 tham gia. R\u00f5 r\u00e0ng vi\u1ec7c tham gia b\u00ean pha \u0111i\u1ec1u ch\u1ec9nh ch\u1ee9a r\u1ea5t nhi\u1ec1u r\u1ee7i ro v\u00ec v\u1eady t\u1ed1t nh\u1ea5t h\u00e3y \u0111\u1ec3 n\u00f3 ho\u00e0n ch\u00ednh s\u1ef1 \u0111i\u1ec1u ch\u1ec9nh sau \u0111\u00f3 x\u00e1c nh\u1eadn vi\u1ec7c \u0111i l\u00ean hay \u0111i xu\u1ed1ng. V\u00ec v\u1eady giao d\u1ecbch theo nguy\u00ean l\u00fd Elliot ch\u1ec9 t\u1eadp trung v\u00e0o s\u00f3ng \u0111\u1ea9y (pha t\u0103ng) v\u00e0 b\u1ecf qua s\u00f3ng \u0111i\u1ec1u ch\u1ec9nh.<\/p>\n

3. Ph\u01b0\u01a1ng ph\u00e1p v\u00e0o l\u1ec7nh trong s\u00f3ng \u0111\u1ea9y<\/strong><\/p>\n

Mua t\u1ea1i s\u00f3ng \u0111\u1ea9y 3:<\/mark><\/strong><\/p>\n

V\u00ec ch\u00fang ta kh\u00f3 ph\u00e1t hi\u1ec7n \u0111\u1ec3 mua h\u00e0ng \u1edf s\u00f3ng 1 n\u00ean ch\u00fang ta s\u1ebd mua \u1edf s\u00f3ng 3. <\/p>\n

C\u00e1c \u0111i\u1ec1u ki\u1ec7n:<\/em><\/strong><\/p>\n

    \n
  • Xu h\u01b0\u1edbng l\u00e0 cp \u0111ang t\u0103ng<\/li>\n
  • S\u00f3ng 1 v\u00e0 2 c\u00f3 \u0111\u1ea9y \u0111\u1ee7 c\u1ea5u tr\u00fac c\u1ee7a m\u1ed9t chu k\u1ef3 nh\u1ecf trong \u0111\u00f3 s\u00f3ng 1 g\u1ed3m 5 s\u00f3ng nh\u1ecf 1,2,3,4,5 v\u00e0 s\u00f3ng 2 g\u1ed3m \u0111\u1ee7 3 s\u00f3ng a,b,c.<\/li>\n
  • C\u00f3 ph\u00e2n k\u1ef3 \u00e2m RSI ho\u1eb7c MACD gi\u1eefa hai s\u00f3ng 3 v\u00e0 5. <\/li>\n<\/ul>\n

    L\u1ec7nh mua:<\/em><\/strong><\/p>\n

      \n
    • K\u00e9o Fibonacci h\u1ed3i quy<\/strong> (tho\u00e1i lui) t\u1eeb \u0111\u1ec9nh xu\u1ed1ng \u0111\u00e1y s\u00f3ng 1<\/li>\n
    • \u0110\u1eb7t l\u1ec7nh mua t\u1ea1i m\u1ee9c Fibonacci 50%. Ho\u1eb7c chia l\u00e0m 2 t\u1ea1i hai m\u1ee9c 50% v\u00e0 61,8%<\/li>\n<\/ul>\n

      D\u1eebng l\u1ed7:<\/em><\/strong><\/p>\n

        \n
      • \u0110\u1eb7t d\u01b0\u1edbi \u0111\u00e1y s\u00f3ng 1<\/li>\n<\/ul>\n

        Ch\u1ed1t l\u1eddi:<\/em><\/strong><\/p>\n

          \n
        • K\u00e9o Fibonacci h\u1ed3i quy <\/strong>cho s\u00f3ng 1. B\u00e1n t\u1ea1i Fibonacci 100% (X<\/strong>) ho\u1eb7c 161,8% (Y<\/strong>) chi\u1ec1u cao c\u1ee7a s\u00f3ng 1.<\/li>\n
        • Khi gi\u00e1 \u0111\u1ea1t X<\/strong> th\u00ec c\u00f3 th\u1ec3 mua m\u1ed9t ph\u1ea7n v\u00e0 k\u00e9o SL t\u1edbi d\u01b0\u1edbi m\u1ee9c X<\/strong> .<\/li>\n
        • Trong qu\u00e1 tr\u00ecnh \u0111i l\u00ean quan s\u00e1t n\u1ebfu c\u00f3 \u0111\u1ee7 5 s\u00f3ng nh\u1ecf ch\u1ea1y t\u1edbi X th\u00ec b\u00e1n to\u00e0n b\u1ed9. N\u1ebfu th\u1ea5y s\u00f3ng ch\u01b0a \u0111i h\u1ebft th\u00ec c\u00f3 th\u1ec3 b\u00e1n 50% v\u00e0 gi\u1eef 50%.<\/li>\n<\/ul>\n

          Mua t\u1ea1i s\u00f3ng \u0111\u1ea9y 5<\/mark><\/strong><\/p>\n

          Gi\u1ea3 s\u1eed ch\u00fang ta kh\u00f4ng t\u1eadn \u0111\u1ee5ng \u0111\u01b0\u1ee3c s\u00f3ng 3 th\u00ec gi\u1edd t\u1edbi s\u00f3ng 5. \u0110i\u1ec1u ki\u1ec7n mua:<\/em><\/strong><\/p>\n

            \n
          • Xu h\u01b0\u1edbng l\u1edbn c\u1ee7a cp v\u1eabn l\u00e0 \u0111ang trong xu h\u01b0\u1edbng t\u0103ng.<\/li>\n
          • S\u00f3ng 1,2,3,4 c\u00f3 \u0111\u1ea7y \u0111\u1ee7 c\u1ea5u tr\u00fac trong \u0111\u00f3 s\u00f3ng 1 v\u00e0 s\u00f3ng 3 g\u1ed3m \u0111\u1ee7 5 s\u00f3ng nh\u1ecf 1,2,3,4,5 v\u00e0 s\u00f3ng 2 v\u00e0 4 c\u00f3 \u0111\u1ee7 s\u00f3ng \u0111i\u1ec1u ch\u1ec9nh a,b,c.<\/li>\n
          • C\u00f3 ph\u00e2n k\u1ef3 \u00e2m RSI ho\u1eb7c MACD gi\u1eefa hai s\u00f3ng 3 v\u00e0 5 c\u1ee7a s\u00f3ng 3<\/li>\n<\/ul>\n

            L\u1ec7nh mua:<\/em><\/strong><\/p>\n

              \n
            • K\u00e9o Fibonacci h\u1ed3i quy t\u1eeb \u0111\u00ecnh xu\u1ed1ng \u0111\u00e1y s\u00f3ng 3<\/li>\n
            • \u0110\u1eb7t l\u1ec7nh mua t\u1ea1i m\u1ee9c FIbonacci 39,2%. N\u1ebfu s\u00f3ng 3 bao g\u1ed3m s\u00f3ng con th\u00ec \u0111\u1eb7t l\u1ec7nh mua t\u1ea1i m\u1ee9c Fibonacci 23,6% c\u1ee7a s\u00f3ng 3.<\/li>\n<\/ul>\n

              L\u1ec7nh d\u1eebng l\u1ed7:<\/strong><\/em><\/p>\n

                \n
              • \u0110\u1eb7t t\u1ea1i m\u1ee9c Fibonacci 50% c\u1ee7a s\u00f3ng 3<\/li>\n<\/ul>\n

                Ch\u1ed1t l\u1eddi:<\/em><\/strong><\/p>\n

                  \n
                • N\u1ebfu s\u00f3ng 1,3 l\u00e0 s\u00f3ng c\u00f3 chi\u1ec1u cao v\u00e0 c\u1ea5u tr\u00fac b\u00ecnh th\u01b0\u1eddng th\u00ec kh\u1ea3 n\u0103ng r\u1ea5t cao s\u00f3ng 5 s\u1ebd t\u0103ng m\u1ea1nh m\u1ebd. K\u00e9o Fibonacci m\u1edf r\u1ed9ng<\/strong> t\u1eeb \u0111\u00e1y c\u1ee7a ch\u00e2n s\u00f3ng 1 t\u1edbi \u0111\u1ec9nh c\u1ee7a s\u00f3ng 3. S\u00f3ng 5 m\u1ee5c ti\u00eau s\u1ebd l\u00e0 161,8% chi\u1ec1u cao c\u1ee7a s\u00f3ng 4 ho\u1eb7c 38,2% chi\u1ec1u cao c\u1ee7a c\u1ea3 s\u00f3ng 1 v\u00e0 3.<\/li>\n
                • Trong tr\u01b0\u1eddng h\u1ee3p s\u00f3ng 5 t\u0103ng m\u1ea1nh th\u00ec kh\u1ea3 n\u0103ng \u0111\u1ec9nh s\u00f3ng s\u1ebd l\u00e0 100% v\u00e0 161,8% c\u1ee7a chi\u1ec1u cao c\u1ea3 s\u00f3ng 1 v\u00e0 3. V\u00ec s\u00f3ng 5 l\u00e0 s\u00f3ng cu\u1ed1i n\u00ean c\u1ea7n c\u1ea9n th\u1eadn h\u01a1n tr\u00e1nh m\u1ea5t h\u1ebft l\u1ee3i nhu\u1eadn ho\u1eb7c l\u1ed7 khi CP b\u01b0\u1edbc v\u00e0o pha \u0111i\u1ec1u ch\u1ec9nh.<\/li>\n<\/ul>\n

                  Fibonacci h\u1ed3i quy hay c\u00f2n g\u1ecdi l\u00e0 tho\u00e1i lui v\u00e0 Fibonacci m\u1edf r\u1ed9ng d\u1ef1a tr\u00ean xu h\u01b0\u1edbng c\u00f3 trong thanh c\u00f4ng c\u1ee5 trong ph\u00e2n t\u00edch k\u1ef9 thu\u1eadt c\u1ee7a c\u00e1c c\u00f4ng ty CK. <\/p>\n

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                  Th\u1ef1c h\u00e0nh ph\u00e2n t\u00edch Fibonacci<\/a><\/p><\/blockquote>\n