{"id":17481,"date":"2024-04-21T05:16:08","date_gmt":"2024-04-21T05:16:08","guid":{"rendered":"https:\/\/soicau4091.minhngocxoso.com\/cach-choi-lo-cap-nuoi-xo-so-mien-bac\/"},"modified":"2024-04-21T05:16:08","modified_gmt":"2024-04-21T05:16:08","slug":"chia-se-kinh-nghiem-nuoi-lo-cap-3-ngay","status":"publish","type":"post","link":"https:\/\/xien2depnhat.com\/chia-se-kinh-nghiem-nuoi-lo-cap-3-ngay\/","title":{"rendered":"chia s\u1ebb kinh nghi\u1ec7m nu\u00f4i l\u00f4 c\u1eb7p 3 ng\u00e0y"},"content":{"rendered":"
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I, CH\u01a0I L\u00d4\u00a0C\u1eb6P NU\u00d4I X\u1ed4 S\u1ed0 MI\u1ec0N B\u1eaeC ?<\/strong><\/h4>\n

Ph\u01b0\u01a1ng ph\u00e1p ch\u01a1i l\u00f4 c\u1eb7p nu\u00f4i \u0111\u01b0\u1ee3c gi\u1edbi ch\u01a1i l\u00f4 \u0111\u1ec1 x\u1ed5 s\u1ed1 mi\u1ec1n b\u1eafc l\u00e0 m\u1ed9t ph\u01b0\u01a1ng ph\u00e1p ch\u01a1i l\u00f4 r\u1ea5t hi\u1ec7u qu\u1ea3 trong b\u1ed9 m\u00f4n s\u1ed1 h\u1ecdc n\u00e0y. \u0110\u00e2y l\u00e0 ph\u01b0\u01a1ng ph\u00e1p mang l\u1ea1i s\u1ef1 \u1ed5n \u0111\u1ecbnh, r\u1ee7i ro th\u1ea5p, kh\u1ea3 n\u0103ng sinh l\u1ee3i cao v\u00ec t\u1ef7 l\u1ec7 n\u1ed5 2 nh\u00e1y r\u1ea5t nhi\u1ec1u.<\/p>\n

1.L\u00f4 c\u1eb7p th\u01b0\u1eddng: C\u00f3\u00a0d\u1ea1ng 00-55, 11-66, 12-21 .(\u0111\u00e1nh l\u00f4 ng\u00e0y th\u00ec n\u00ean ch\u1ecdn l\u00f4 c\u1eb7p th\u01b0\u1eddng tr\u00e1nh tr\u01b0\u1eddng h\u1ee3p v\u1ec1 l\u1ed9n)<\/p>\n

2.L\u00f4 c\u1eb7p t\u1ef1 do: C\u00f3 d\u1ea1ng 11-52, 12-46,65-43 \u2026 l\u00e0 nh\u1eefng c\u1eb7p l\u00f4 t\u00f4 b\u1ea5t k\u1ef3 g\u00e9p l\u1ea1i v\u1edbi nhau<\/p>\n

3.L\u00f4 c\u1eb7p tr\u00f9ng \u0111\u1ea7u: C\u00f3 d\u1ea1ng 12-14, 62-64\u2026 l\u00e0 nh\u1eefng c\u1eb7p l\u00f4 c\u00f3 \u0111\u1ea7u gi\u1ed1ng nhau\u00a0(\u00e1p d\u1ee5ng cho ki\u1ec3u nu\u00f4i 2-3 ng\u00e0y r\u1ea5t hi\u1ec7u qu\u1ea3)<\/p>\n

4.L\u00f4 c\u1eb7p tr\u00f9ng \u0111u\u00f4i : C\u00f3 d\u1ea1ng 56-76, 13-23, 71-81\u00a0\u2026 l\u00e0 nh\u1eefng c\u1eb7p l\u00f4 c\u00f3 \u0111u\u00f4i\u00a0gi\u1ed1ng nhau (\u00e1p d\u1ee5ng cho ki\u1ec3u nu\u00f4i 2-3 ng\u00e0y r\u1ea5t hi\u1ec7u qu\u1ea3<\/p>\n

\u0110\u00e1nh l\u00f4 c\u1eb7p nu\u00f4i 2 ng\u00e0y c\u0169ng \u0111\u01b0\u1ee3c, 3 ng\u00e0y c\u0169ng \u0111\u01b0\u1ee3c, t\u00f9y v\u00e0o kh\u1ea3 n\u0103ng t\u00e0i ch\u00ednh v\u00e0 c\u00e1ch ch\u1ecdn khung c\u1ee7a b\u1ea1n. Nh\u01b0ng theo kinh nghi\u1ec7m c\u1ee7a ch\u00fang t\u00f4i th\u00ec ph\u01b0\u01a1ng ph\u00e1p l\u00f4 c\u1eb7p tr\u00f9ng \u0111\u1ea7u ho\u1eb7c tr\u00f9ng \u0111u\u00f4i nu\u00f4i 3 ng\u00e0y kh\u00e1 l\u00e0 hi\u1ec7u qu\u1ea3 v\u00e0 n\u1ed5 2-3 nh\u00e1y r\u1ea5t nhi\u1ec1u<\/p>\n

II,\u00a0C\u00c1CH CH\u01a0I \u201cL\u00d4\u00a0C\u1eb6P NU\u00d4I 3 NG\u00c0Y\u201d<\/strong><\/h4>\n

L\u00f4 c\u1eb7p nu\u00f4i 3 ng\u00e0y\u00a0l\u00e0 ph\u01b0\u01a1ng ph\u00e1p ch\u01a1i \u0111\u01b0\u1ee3c ti\u1ebfn h\u00e0nh nh\u01b0 sau:\u00a0B\u1ea1n ch\u1ecdn m\u1ed9t\u00a0c\u1eb7p l\u00f4 t\u00f4\u00a0n\u00e0o \u0111\u00f3 m\u00e0 b\u1ea1n cho r\u1eb1ng trong 3 ng\u00e0y t\u1edbi n\u00f3 nh\u1ea5t \u0111\u1ecbnh s\u1ebd v\u1ec1 \u00edt nh\u1ea5t 1 trong 2 con : V\u00ed d\u1ee5\u00a0b\u1ea1n ch\u1ecdn c\u1eb7p l\u00f4 tr\u00f9ng \u0111\u1ea7u\u00a014-16 v\u00e0 b\u1ea1n cho r\u1eb1ng trong ng\u00e0y h\u00f4m nay, ng\u00e0y mai ho\u1eb7c \u00edt nh\u1ea5t l\u00e0 ng\u00e0y kia n\u00f3 s\u1ebd v\u1ec1 1 trong 2 con. C\u00e1c b\u1ea1n s\u1ebd v\u00e0o ti\u1ec1n cho t\u1eebng con v\u1edbi t\u1ef7 l\u1ec7\u00a01 : 3.5 : 9. V\u00ed d\u1ee5 ng\u00e0y 1 b\u1ea1n b\u1eaft \u0111\u1ea7u \u0111\u00e1nh c\u1eb7p 14-16 m\u1ed7i con 1 \u0111i\u1ec3m\u00a0th\u00ec s\u1ed1 \u0111i\u1ec3m c\u1ea7n \u0111\u1ec3 \u0111\u00e1nh cho c\u1eb7p \u0111\u00f3 trong c\u00e1c ng\u00e0y l\u00e0<\/p>\n

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  1. S\u1ed1 \u0111i\u1ec3m \u0111\u00e1nh c\u1ee7a ng\u00e0y 1 \u00a0= 1 \u0111i\u1ec3m *\u00a01 x 2 \u00a0 \u00a0= \u00a02\u00a0\u0111i\u1ec3m m\u1ed9t c\u1eb7p, 1 \u0111i\u1ec3m m\u1ed7i con<\/li>\n
  2. S\u1ed1 \u0111i\u1ec3m \u0111\u00e1nh c\u1ee7a ng\u00e0y 2 \u00a0 = 1 \u0111i\u1ec3m *\u00a03.5\u00a0x 2 = \u00a07\u00a0\u0111i\u1ec3m m\u1ed9t c\u1eb7p, 3.5 \u0111i\u1ec3m m\u1ed7i con<\/li>\n
  3. S\u1ed1 \u0111i\u1ec3m \u0111\u00e1nh c\u1ee7a ng\u00e0y 3\u00a0 \u00a0= 1 \u0111i\u1ec3m * 9\u00a0x 2 \u00a0 \u00a0= \u00a018\u00a0\u0111i\u1ec3m m\u1ed9t c\u1eb7p, 9 \u0111i\u1ec3m m\u1ed7i con<\/li>\n<\/ol>\n