{"id":17860,"date":"2024-03-24T11:42:14","date_gmt":"2024-03-24T11:42:14","guid":{"rendered":"https:\/\/soicau3075.minhngocxoso.com\/tong-hop-nhung-phuong-phap-soi-cau-lo-de-chuan-tu-chuyen-gia-hdxs\/"},"modified":"2024-07-06T08:32:00","modified_gmt":"2024-07-06T01:32:00","slug":"tong-hop-nhung-phuong-phap-soi-cau-lo-de-chuan-tu-chuyen-gia-hdxs","status":"publish","type":"post","link":"https:\/\/dudoanxsmbdep.com\/tong-hop-nhung-phuong-phap-soi-cau-lo-de-chuan-tu-chuyen-gia-hdxs\/","title":{"rendered":"T\u1ed5ng h\u1ee3p nh\u1eefng ph\u01b0\u01a1ng ph\u00e1p soi c\u1ea7u l\u00f4 \u0111\u1ec1 chu\u1ea9n t\u1eeb chuy\u00ean gia H\u0110XS"},"content":{"rendered":"
D\u01b0\u1edbi \u0111\u00e2y l\u00e0 nh\u1eefng ph\u01b0\u01a1ng ph\u00e1p soi c\u1ea7u l\u00f4 \u0111\u1ec1<\/strong>, chia s\u1ebb kinh nghi\u1ec7m soi c\u1ea7u l\u00f4 \u0111\u1ec1<\/em> hi\u1ec7u qu\u1ea3 cao t\u1eeb c\u00e1c chuy\u00ean gia H\u0110XS mi\u1ec1n b\u1eafc. C\u00e1ch b\u1eaft c\u1ea7u sao cho chu\u1ea9n \u00edt b\u1ecb r\u1ee7i ro nh\u1ea5t, gi\u00fap anh ch\u1ecb em xa b\u1edd c\u00f3 nhi\u1ec1u c\u01a1 h\u1ed9i chi\u1ebfn th\u1eafng h\u01a1n. Tuy nhi\u00ean m\u1ecdi ng\u01b0\u1eddi ch\u1ec9\u00a0 \u0111\u1ecdc \u0111\u1ec3 tham kh\u1ea3o r\u1ed3i r\u00fat ra k\u1ebft lu\u1eadn cho b\u1ea3n th\u00e2n ch\u1ee9 kh\u00f4ng n\u00ean \u00e1p d\u1ee5ng 1 c\u00e1ch kh\u00f4ng c\u00f3 khoa h\u1ecdc v\u00e0 thi\u1ebfu c\u0103n c\u1ee9.<\/p>\n C\u00e1ch soi c\u1ea7u n\u00e0y d\u1ef1a v\u00e0o gi\u1ea3i \u0111\u1eb7c bi\u1ec7t c\u1ee7a 2 h\u00f4m li\u00ean ti\u1ebfp \u0111\u1ec3 \u0111\u00e1nh h\u00f4m th\u1ee9 3. Ta x\u00e9t c\u00e1c tr\u01b0\u1eddng h\u1ee3p sau \u0111\u1ec3\u00a0soi c\u1ea7u l\u00f4 ch\u00ednh x\u00e1c 100%:<\/strong><\/p>\n Tr\u01b0\u1eddng h\u1ee3p 1:<\/strong>\u00a0M < 10, N < 10. Ta l\u1ea5y hi\u1ec7u 10 \u2013 M = X, 10 \u2013 N = Y. Ta s\u1ebd soi c\u1ea7u l\u00f4 ch\u00ednh x\u00e1c 100% ng\u00e0y th\u1ee9 3 l\u00e0 XY.<\/p>\n Tr\u01b0\u1eddng h\u1ee3p 2:<\/strong>\u00a0M > 10, N < 10. Ta l\u1ea5y hi\u1ec7u 20 \u2013 M = X, 10 \u2013 N = Y. Ta s\u1ebd soi c\u1ea7u l\u00f4 ch\u00ednh x\u00e1c 100% ng\u00e0y th\u1ee9 3 l\u00e0 XY.<\/p>\n Tr\u01b0\u1eddng h\u1ee3p 3:<\/strong>\u00a0M > 10, N > 10. Ta l\u1ea5y hi\u1ec7u 20 \u2013 M = X, 20 \u2013 N = Y. Ta s\u1ebd soi c\u1ea7u l\u00f4 ch\u00ednh x\u00e1c 100% ng\u00e0y th\u1ee9 3 l\u00e0 XY.<\/p>\n Tr\u01b0\u1eddng h\u1ee3p 4:<\/strong>\u00a0M < 10, N > 10. Ta l\u1ea5y hi\u1ec7u 10 \u2013 M = X, 20 \u2013 N = Y. Ta s\u1ebd soi c\u1ea7u l\u00f4 ch\u00ednh x\u00e1c 100% ng\u00e0y th\u1ee9 3 l\u00e0 YX (tr\u01b0\u1eddng h\u1ee3p \u0111\u1eb7c bi\u1ec7t).<\/p>\n L\u01b0u \u00fd:\u00a0<\/strong>N\u1ebfu t\u1ed5ng M, N l\u00e0 0 ho\u1eb7c 10 th\u00ec X, Y \u0111\u1ec1u l\u00e0 0 h\u1ebft.<\/p>\n V\u00ed d\u1ee5 1:<\/strong>\u00a0\u0110\u1ec1 h\u00f4m th\u1ee9 nh\u1ea5t v\u1ec1 36, \u0111\u1ec1 h\u00f4m th\u1ee9 2 v\u1ec1 93. V\u00ed d\u1ee5 2:<\/strong>\u00a0\u0110\u1ec1 h\u00f4m th\u1ee9 nh\u1ea5t v\u1ec1 74, \u0111\u1ec1 h\u00f4m th\u1ee9 2 v\u1ec1 85. Tr\u00ean \u0111\u00e2y l\u00e0 l\u00fd thuy\u1ebft c\u1ee7a c\u00e1ch soi c\u1ea7u l\u00f4 m\u1edbi nh\u1ea5t 2024 ch\u00ednh x\u00e1c 100%. Sau \u0111\u00e2y m\u00ecnh l\u1ea5y v\u00ed d\u1ee5 th\u1ef1c t\u1ebf cho c\u00e1c b\u1ea1n d\u1ec5 hi\u1ec3u v\u00e0 s\u1ef1 ch\u00ednh x\u00e1c c\u1ee7a n\u00f3.<\/p>\n <\/p>\n M\u00ecnh xin l\u1ea5y k\u1ebft qu\u1ea3 c\u1ee7a th\u00e1ng 1\/2024 l\u00e0m v\u00ed d\u1ee5 th\u1ef1c t\u1ebf c\u00e1ch v\u00e0o ti\u1ec1n c\u1ee7a m\u00ecnh cho c\u00e1c b\u1ea1n xem c\u00e1ch\u00a0soi c\u1ea7u l\u00f4 ch\u00ednh x\u00e1c 100%\u00a0<\/strong>hi\u1ec7u qu\u1ea3 nh\u01b0 th\u1ebf n\u00e0o:<\/p>\n \u0110\u00e2y l\u00e0 m\u1ed9t c\u00e1ch t\u00ednh l\u00f4 \u0111\u1ec1 r\u1ea5t m\u1edbi v\u00e0 hi\u1ec7u qu\u1ea3. C\u00e1ch t\u00ednh n\u00e0y \u0111\u01b0\u1ee3c s\u1eed d\u1ee5ng khi 3 ng\u00e0y kh\u00f4ng c\u00f3 l\u00f4 r\u01a1i (l\u00f4 r\u01a1i l\u00e0 g\u00ec c\u00f3 th\u1ec3 tham kh\u1ea3o b\u00e0i vi\u1ebft v\u1ec1\u00a0l\u00f4 r\u01a1i<\/strong>) r\u1ed3i l\u1ea5y s\u1ed1 l\u1ed9n c\u1ee7a \u0111\u1ec1 3 h\u00f4m tr\u01b0\u1edbc l\u00e0 l\u00f4 b\u1ea1ch th\u1ee7 c\u1ee7a ng\u00e0y h\u00f4m nay.<\/p>\n \u2013 \u0110\u1ec3 \u00fd xem 3 ng\u00e0y li\u00ean t\u1ee5c l\u00f4 kh\u00f4ng r\u01a1i.<\/p>\n \u2013 Ta l\u1ea5y s\u1ed1 l\u1ed9n c\u1ee7a \u0111\u1ec1 3 ng\u00e0y tr\u01b0\u1edbc (V\u00ed d\u1ee5 h\u00f4m nay 27\/02 th\u00ec l\u1ea5y ng\u00e0y 24\/2).<\/p>\n \u2013 X\u00e9t s\u1ed1 l\u1ed9n ta t\u00ecm \u0111\u01b0\u1ee3c xem 3 ng\u00e0y li\u00ean t\u1ee5c tr\u01b0\u1edbc, n\u1ebfu l\u00f4 c\u1ee7a n\u00f3 kh\u00f4ng v\u1ec1 th\u00ec s\u1ebd l\u00e0m b\u1ea1ch th\u1ee7 l\u00f4 \u0111\u00e1nh ng\u00e0y h\u00f4m nay.<\/p>\n C\u00e1ch t\u00ednh l\u00f4 \u0111\u1ec1 n\u00e0y d\u1ef1a v\u00e0o quy lu\u1eadt l\u00f4 hay v\u1ec1 c\u1ee7a 2024, b\u1ea1n n\u00ean t\u00ecm ra 2 l\u00f4 hay v\u1ec1 nh\u1ea5t c\u1ee7a n\u0103m nay, v\u00e0 b\u1ea1n th\u00eam s\u1ed1 l\u1ed9n c\u1ee7a 2 l\u00f4 \u0111\u00f3, \u0111\u1ec3 \u0111\u00e1nh theo khung max 3 ng\u00e0y. Ti\u1ec1n v\u1ec1 r\u1ea5t \u1ed5n \u0111\u1ecbnh.<\/p>\n \u2013 T\u00ecm ra l\u00f4 hay v\u1ec1 nh\u1ea5t trong n\u0103m, th\u00eam 2 s\u1ed1 l\u1ed9n c\u1ee7a n\u00f3 n\u1eefa, l\u00e0 b\u1ea1n \u0111\u01b0\u1ee3c 2 c\u1eb7p l\u00f4 hay v\u1ec1 c\u1ee7a n\u0103m nay.<\/p>\n \u2013 Xem th\u1ed1ng k\u00ea n\u1ebfu \u0111\u1ebfn chu k\u1ef3 n\u00f3 v\u1ec1 th\u00ec c\u1eb7p l\u00f4 \u0111\u00f3 s\u1ebd v\u1ec1 li\u00ean t\u1ee5c.<\/p>\n \u2013 Khi n\u00e0o 3,4 ng\u00e0y kh\u00f4ng v\u1ec1 ho\u1eb7c th\u1ea5y chu k\u1ef3 n\u00f3 v\u1ec1 nhi\u1ec1u th\u00ec \u0111\u00e1nh c\u1ea3 c\u1eb7p trong khung 3 ng\u00e0y, t\u1ea5t nhi\u00ean s\u1ebd c\u00f3 l\u00e3i.<\/p>\n C\u00e1ch soi c\u1ea7u x\u1ed5 s\u1ed1 mi\u1ec1n b\u1eafc n\u00e0y l\u00e0 m\u1ed9t s\u1ef1 tr\u00f9ng h\u1ee3p trong c\u00e1ch soi c\u1ea7u k\u1ebft qu\u1ea3 ng\u00e0y h\u00f4m tr\u01b0\u1edbc v\u00e0 k\u1ebft qu\u1ea3 ng\u00e0y h\u00f4m sau. Theo c\u00e1ch soi c\u1ea7u n\u00e0y th\u00ec c\u00e1c b\u1ea1n ph\u1ea3i \u0111\u00e1nh 2 con l\u00f4 trong 3 ng\u00e0y khi n\u00e0o tr\u00fang th\u00ec chuy\u1ec3n sang \u0111\u00e1nh con kh\u00e1c, n\u00ean c\u00e1c b\u1ea1n ph\u1ea3i chu\u1ea9n b\u1ecb m\u1ed9t s\u1ed1 v\u1ed1n l\u1edbn \u0111\u1ec3 theo c\u00e1ch soi c\u1ea7u n\u00e0y. Nh\u01b0ng ng\u01b0\u1ee3c l\u1ea1i n\u00f3 c\u00f3 t\u00ednh an to\u00e0n cao v\u00e0 mang l\u1ea1i t\u00e2m l\u00fd tho\u1ea3i m\u00e1i v\u00e0 ch\u00ednh x\u00e1c nh\u1ea5t cho ng\u01b0\u1eddi soi c\u1ea7u x\u1ed5 s\u1ed1.<\/p>\n B\u01b0\u1edbc 1:<\/b>\u00a0L\u1ea5y t\u1ed5ng s\u1ed1 \u0111\u1ea7u v\u00e0 s\u1ed1 cu\u1ed1i c\u1ee7a gi\u1ea3i \u0111\u1eb7c bi\u1ec7t ng\u00e0y h\u00f4m nay. N\u1ebfu t\u1ed5ng \u0111\u00f3 > 10 th\u00ec ta c\u1ed9ng ti\u1ebfp t\u1eebng s\u1ed1 trong t\u1ed5ng \u0111\u00f3, c\u00f2n n\u1ebfu = 10 th\u00ec ra l\u1ea5y b\u1eb1ng 0. N\u1ebfu < 10 th\u00ec l\u1ea5y lu\u00f4n t\u1ed5ng \u0111\u00f3. B\u01b0\u1edbc 2:<\/b>\u00a0L\u1ea5y b\u00f3ng c\u1ee7a t\u1ed5ng \u0111\u00f3 ta \u0111\u01b0\u1ee3c m\u1ed9t s\u1ed1 g\u1ecdi l\u00e0 X. (n\u1ebfu ch\u01b0a bi\u1ebft b\u00f3ng l\u00f4 \u0111\u1ec1 c\u00f3 th\u1ec3 t\u00ecm hi\u1ec3u \u1edf \u0111\u00e2y:\u00a0b\u00f3ng l\u00f4 \u0111\u1ec1<\/strong> B\u01b0\u1edbc 3:<\/b>\u00a0Ta th\u1ed1ng k\u00ea gi\u1ea3i b\u1ea3y trong k\u1ebft qu\u1ea3 x\u1ed5 s\u1ed1 ta s\u1ebd c\u00f3 4 gi\u1ea3i, m\u1ed7i gi\u1ea3i c\u00f3 2 ch\u1eef s\u1ed1 n\u00ean ta s\u1ebd c\u00f3 \u0111\u01b0\u1ee3c 8 ch\u1eef s\u1ed1. Ti\u1ebfp theo ch\u1ecdn 2 s\u1ed1 xu\u1ea5t hi\u1ec7n nhi\u1ec1u nh\u1ea5t trong 8 s\u1ed1 \u0111\u00f3, n\u1ebfu nhi\u1ec1u s\u1ed1 xu\u1ea5t hi\u1ec7n b\u1eb1ng nhau ta \u01b0u ti\u00ean s\u1ed1 xu\u1ea5t hi\u1ec7n tr\u01b0\u1edbc t\u1eeb tr\u00e1i qua ph\u1ea3i theo k\u1ebft qu\u1ea3 x\u1ed5 s\u1ed1. Ta \u0111\u01b0\u1ee3c 2 s\u1ed1 l\u00e0 A,B. B\u01b0\u1edbc 4:<\/b>\u00a0Ta gh\u00e9p X \u0111\u00e3 t\u00ecm \u0111\u01b0\u1ee3c \u1edf b\u01b0\u1edbc 2 v\u1edbi hai s\u1ed1 t\u00ecm \u0111\u01b0\u1ee3c \u1edf b\u01b0\u1edbc 3, ta s\u1ebd \u0111\u01b0\u1ee3c 2 l\u00f4 l\u00e0 XA v\u00e0 XB. B\u01b0\u1edbc 5:<\/b>\u00a0\u0110\u00e1nh 2 l\u00f4 t\u00ecm \u0111\u01b0\u1ee3c trong 3 ng\u00e0y, n\u1ebfu tr\u00fang th\u00ec b\u1ecf 2 l\u00f4 \u0111\u00f3, t\u00ecm 2 l\u00f4 kh\u00e1c \u0111\u00e1nh. N\u1ebfu tr\u01b0\u1ee3t th\u00ec ng\u00e0y sau \u0111\u00e1nh g\u1ea5p \u0111\u00f4i ti\u1ec1n ng\u00e0y tr\u01b0\u1edbc. N\u1ebfu 3 ng\u00e0y kh\u00f4ng v\u1ec1 th\u00ec b\u1ecf, nu\u00f4i ti\u1ebfp 3 ng\u00e0y kh\u00e1c. N\u1ebfu \u0111\u00e1nh 1 tu\u1ea7n m\u00e0 c\u00f3 l\u00e3i th\u00ec ngh\u1ec9 1 tu\u1ea7n r\u1ed3i \u0111\u00e1nh ti\u1ebfp, c\u00f2n kh\u00f4ng c\u00f3 l\u00e3i th\u00ec \u0111\u00e1nh ti\u1ebfp tu\u1ea7n ti\u1ebfp theo c\u00f3 l\u00e3i r\u1ed3i ngh\u1ec9 1 tu\u1ea7n. Nh\u1eefng c\u00e1ch t\u00ednh l\u00f4 \u0111\u1ec1 tr\u00ean \u0111\u00e3 \u0111\u01b0\u1ee3c ki\u1ec3m nghi\u1ec7m l\u00e0 nh\u1eefng\u00a0c\u00e1ch t\u00ednh l\u00f4 \u0111\u1ec1 hi\u1ec7u qu\u1ea3 nh\u1ea5t<\/strong>. Chu k\u1ef3 r\u1ea5t \u1ed5n \u0111\u1ecbnh, r\u1ea5t \u00edt khi ch\u1ec7ch v\u00e0 khan. N\u00ean nh\u1eefng ai theo c\u00e1c c\u00e1ch tr\u00ean lu\u00f4n lu\u00f4n an t\u00e2m v\u00e0 \u0111\u1ea1t hi\u1ec3u qu\u1ea3 cao nh\u1ea5<\/strong>t.\u00a0Nh\u01b0 v\u1eady \u0111\u00e2y l\u00e0 m\u1ed9t s\u1ed1 ph\u01b0\u01a1ng ph\u00e1p\u00a0soi c\u1ea7u x\u1ed5 s\u1ed1 mi\u1ec1n b\u1eafc ch\u00ednh x\u00e1c nh\u1ea5t<\/b> hi\u1ec7n t\u1ea1i . V\u00ec n\u00f3 lu\u00f4n mang l\u1ea1i l\u1ee3i nhu\u1eadn cho c\u00e1c b\u1ea1n, ch\u00fac c\u00e1c b\u1ea1n may m\u1eafn v\u00e0 soi c\u1ea7u th\u1eadt ch\u00ednh x\u00e1c.<\/p>\nNguy\u00ean l\u00fd soi c\u1ea7u l\u00f4 ch\u00ednh x\u00e1c 100%<\/span><\/h2>\n
\nV\u00ed d\u1ee5: \u0110\u1ec1 h\u00f4m th\u1ee9 nh\u1ea5t l\u00e0 AB, \u0110\u1ec1 h\u00f4m th\u1ee9 2 l\u00e0 CD.
\nTa x\u00e9t t\u1ed5ng \u0111\u1ec1 h\u00f4m th\u1ee9 nh\u1ea5t: A + B = M.
\nT\u1ed5ng \u0111\u1ec1 h\u00f4m th\u1ee9 2: C + D = N.<\/p>\n
\nT\u1eeb \u0111\u1ec1 hai h\u00f4m li\u00ean ti\u1ebfp ta t\u00ecm \u0111\u01b0\u1ee3c X = 10 -(3+6) =1. Y = 20 \u2013 (9+3) = 8.
\nTa th\u1ea5y \u0111\u00e2y thu\u1ed9c tr\u01b0\u1eddng h\u1ee3p 4 n\u00ean con l\u00f4 ng\u00e0y th\u1ee9 3 ta \u0111\u00e1nh 81.<\/p>\n
\nT\u1eeb k\u1ebft qu\u1ea3 \u0111\u1ec1 tr\u00ean ta t\u00ecm \u0111\u01b0\u1ee3c X = 20 \u2013 (7+4)= 9. Y = 20 \u2013 (8+5)= 7.
\nN\u00ean ta t\u00ecm \u0111\u01b0\u1ee3c XY l\u00e0 97 \u0111\u1ec3 \u0111\u00e1nh l\u00f4 ng\u00e0y th\u1ee9 3.<\/p>\nKhi n\u00e0o th\u00ec d\u00f9ng c\u00e1ch n\u00e0y \u0111\u1ec3 soi c\u1ea7u l\u00f4 hi\u1ec7u qu\u1ea3:<\/span><\/h3>\n
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\n=> C\u00e1c b\u1ea1n c\u1ea3m th\u1ea5y khi 5, 6 ng\u00e0y c\u1ea7u n\u00e0y kh\u00f4ng ra, ta b\u1eaft \u0111\u1ea7u \u0111\u00e1nh, n\u1ebfu kh\u00f4ng v\u1ec1 ngh\u1ec9 m\u1ed9t h\u00f4m r\u1ed3i \u0111\u00e1nh ti\u1ebfp 3 ng\u00e0y, ho\u1eb7c th\u1ea5y c\u1ea7u n\u00e0y ra \u0111\u00e1nh ti\u1ebfp h\u00f4m sau.<\/li>\n<\/ol>\nV\u00ed d\u1ee5 th\u1ef1c t\u1ebf soi c\u1ea7u l\u00f4 ch\u00ednh x\u00e1c 100%<\/span><\/h3>\n
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C\u00e1c ph\u01b0\u01a1ng ph\u00e1p t\u00ednh l\u00f4 \u0111\u1ec1 hi\u1ec7u qu\u1ea3 t\u1eeb chuy\u00ean gia H\u0110XS mi\u1ec1n b\u1eafc<\/span><\/h2>\n
1.Ph\u01b0\u01a1ng ph\u00e1p t\u00ednh l\u00f4 r\u01a1i l\u1ed9n c\u1ee7a 3 h\u00f4m tr\u01b0\u1edbc.<\/h3>\n
2. C\u00e1ch t\u00ednh l\u00f4 \u0111\u1ec1 hi\u1ec7u qu\u1ea3 d\u1ef1a v\u00e0o l\u00f4 c\u00f3 chu k\u1ef3 hay v\u1ec1.<\/span><\/h3>\n
3. Nguy\u00ean l\u00fd soi c\u1ea7u x\u1ed5 s\u1ed1 mi\u1ec1n b\u1eafc ch\u00ednh x\u00e1c nh\u1ea5t.<\/b><\/h3>\n
\nV\u00ed d\u1ee5:<\/i><\/b>\u00a0Gi\u1ea3i \u0111\u1eb7c bi\u1ec7t ng\u00e0y 1\/9\/2024: 94102 (9 l\u00e0 s\u1ed1 \u0111\u1ea7u, 2 l\u00e0 s\u1ed1 cu\u1ed1i)
\nTa l\u1ea5y t\u1ed5ng = 9+2 = 11. V\u00ec t\u1ed5ng l\u1edbn h\u01a1n 10 n\u00ean ta l\u1ea5y t\u1ed5ng ti\u1ebfp = 1+1 = 2. V\u1eady ta \u0111\u01b0\u1ee3c t\u1ed5ng b\u1eb1ng 2.
\n(gi\u1ea3 s\u1eed n\u1ebfu t\u1ed5ng b\u1eb1ng 10 ta l\u1ea5y l\u00e0 0, c\u00f2n n\u1ebfu t\u1ed5ng b\u1eb1ng s\u1ed1 b\u00e9 h\u01a1n 10 th\u00ec ta l\u1ea5y s\u1ed1 \u0111\u00f3 lu\u00f4n).<\/p>\n
\nV\u00ed d\u1ee5:<\/i><\/b>\u00a0nh\u01b0 t\u1ed5ng \u1edf tr\u00ean b\u1eb1ng 2 th\u00ec b\u00f3ng c\u1ee7a 2 l\u00e0 7. V\u1eady X = 7.<\/p>\n
\nV\u00ed d\u1ee5:<\/i><\/b>\u00a0Gi\u1ea3i b\u1ea3y c\u1ee7a ng\u00e0y 1\/9\/2024 l\u00e0: 73, 37, 03, 55. Ta th\u1ea5y s\u1ed1 3 xu\u1ea5t hi\u1ec7n 3 l\u1ea7n, s\u1ed1 7 xu\u1ea5t hi\u1ec7n 2 l\u1ea7n, s\u1ed1 5 xu\u1ea5t hi\u1ec7n 2 l\u1ea7n.
\nTa l\u1ef1a ch\u1ecdn s\u1ed1 3 v\u00e0 s\u1ed1 7 (lo\u1ea1i s\u1ed1 5 v\u00ec s\u1ed1 7 xu\u1ea5t hi\u1ec7n tr\u01b0\u1edbc t\u1eeb tr\u00e1i sang ph\u1ea3i).
\nV\u1eady ta \u0111\u01b0\u1ee3c s\u1ed1 3 v\u00e0 s\u1ed1 7.<\/p>\n
\nV\u00ed d\u1ee5:<\/i><\/b>\u00a0Nh\u01b0 \u1edf c\u00e1c v\u00ed d\u1ee5 tr\u00ean l\u00e0 c\u00f3 X=7, gh\u00e9p v\u1edbi 2 s\u1ed1 \u1edf l\u00e0 3 v\u00e0 7 t\u00ecm \u0111\u01b0\u1ee3c \u1edf tr\u00ean. Ta s\u1ebd c\u00f3 2 l\u00f4 l\u00e0 73, 77.<\/p>\n
\nV\u00ed d\u1ee5:<\/i><\/b>\u00a0\u0110\u00e1nh l\u00f4 73,77 v\u00e0o ng\u00e0y 2\/9\/2024 \u0111\u00e3 c\u00f3 ngay l\u00f4 73 cho c\u00e1c b\u1ea1n.<\/p>\n