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Least Common Multiple l\u00e0 g\u00ec?<\/b><\/h1>\n
Least Common Multiple (LCM), hay b\u1ed9i chung nh\u1ecf nh\u1ea5t (BCNN), l\u00e0 s\u1ed1 nguy\u00ean d\u01b0\u01a1ng nh\u1ecf nh\u1ea5t chia h\u1ebft cho t\u1ea5t c\u1ea3 c\u00e1c s\u1ed1 \u0111\u00e3 cho.<\/span><\/p>\n V\u00ed d\u1ee5, c\u00e1c b\u1ed9i chung d\u01b0\u01a1ng c\u1ee7a 4 v\u00e0 6 l\u00e0 12, 24, 36, \u2026 Trong \u0111\u00f3, 12 l\u00e0 s\u1ed1 nh\u1ecf nh\u1ea5t. V\u00ec v\u1eady:<\/span><\/p>\n LCM(4,6) = 12<\/span><\/p>\n Trong \u0111\u1ec1 thi SAT Math, c\u00e2u h\u1ecfi th\u01b0\u1eddng \u0111\u01b0\u1ee3c di\u1ec5n \u0111\u1ea1t b\u1eb1ng ti\u1ebfng Anh. Do \u0111\u00f3, h\u1ecdc sinh c\u1ea7n nh\u1eadn di\u1ec7n c\u00e1c c\u1ee5m t\u1eeb nh\u01b0 \u201cleast common multiple\u201d (b\u1ed9i chung nh\u1ecf nh\u1ea5t), \u201csmallest positive integer divisible by\u2026\u201d (s\u1ed1 nguy\u00ean d\u01b0\u01a1ng nh\u1ecf nh\u1ea5t chia h\u1ebft cho\u2026) ho\u1eb7c \u201cmultiple of both\u201d (b\u1ed9i chung c\u1ee7a c\u1ea3 hai s\u1ed1). \u0110\u00e2y l\u00e0 nh\u1eefng d\u1ea5u hi\u1ec7u quan tr\u1ecdng gi\u00fap x\u00e1c \u0111\u1ecbnh khi n\u00e0o c\u1ea7n \u00e1p d\u1ee5ng ki\u1ebfn th\u1ee9c LCM \u0111\u1ec3 gi\u1ea3i b\u00e0i to\u00e1n ch\u00ednh x\u00e1c v\u00e0 hi\u1ec7u qu\u1ea3 h\u01a1n.<\/span><\/p>\n Theo \u0111\u1eb7c t\u1ea3 c\u1ee7a College Board, ph\u1ea7n SAT Math \u0111\u01b0\u1ee3c chia th\u00e0nh b\u1ed1n m\u1ea3ng n\u1ed9i dung ch\u00ednh: Algebra, Advanced Math, Problem-Solving and Data Analysis, v\u00e0 Geometry and Trigonometry. LCM (b\u1ed9i chung nh\u1ecf nh\u1ea5t) kh\u00f4ng \u0111\u01b0\u1ee3c t\u00e1ch th\u00e0nh m\u1ed9t ch\u1ee7 \u0111\u1ec1 \u0111\u1ed9c l\u1eadp, m\u00e0 th\u01b0\u1eddng \u0111\u01b0\u1ee3c v\u1eadn d\u1ee5ng nh\u01b0 m\u1ed9t k\u1ef9 n\u0103ng n\u1ec1n t\u1ea3ng trong nhi\u1ec1u d\u1ea1ng b\u00e0i to\u00e1n, \u0111\u1eb7c bi\u1ec7t l\u00e0:<\/span><\/p>\n Nh\u00ecn chung, c\u00e1c b\u00e0i to\u00e1n li\u00ean quan \u0111\u1ebfn LCM trong SAT Math th\u01b0\u1eddng kh\u00f4ng qu\u00e1 ph\u1ee9c t\u1ea1p v\u1ec1 m\u1eb7t t\u00ednh to\u00e1n. Th\u1eed th\u00e1ch l\u1edbn h\u01a1n n\u1eb1m \u1edf vi\u1ec7c nh\u1eadn di\u1ec7n \u0111\u00fang t\u00ecnh hu\u1ed1ng c\u1ea7n s\u1eed d\u1ee5ng LCM v\u00e0 l\u1ef1a ch\u1ecdn ph\u01b0\u01a1ng ph\u00e1p gi\u1ea3i ph\u00f9 h\u1ee3p, thay v\u00ec \u00e1p d\u1ee5ng c\u00f4ng th\u1ee9c m\u1ed9t c\u00e1ch m\u00e1y m\u00f3c.<\/span><\/p>\n Nh\u00e2n t\u1ea5t c\u1ea3 c\u00e1c s\u1ed1 l\u1ea1i v\u1edbi nhau: 12 \u00d7 18 = 216 l\u00e0 m\u1ed9t b\u1ed9i chung, nh\u01b0ng kh\u00f4ng ph\u1ea3i b\u1ed9i chung nh\u1ecf nh\u1ea5t. \u0110\u00e1p \u00e1n 216 th\u01b0\u1eddng \u0111\u01b0\u1ee3c \u0111\u1eb7t s\u1eb5n l\u00e0m ph\u01b0\u01a1ng \u00e1n nhi\u1ec5u.<\/span><\/p>\n Nh\u1ea7m LCM v\u1edbi GCD: \u0111\u1ec1 h\u1ecfi “greatest common factor” m\u00e0 t\u00ednh LCM, ho\u1eb7c ng\u01b0\u1ee3c l\u1ea1i. M\u1eb9o nh\u1edb: LCM lu\u00f4n l\u1edbn h\u01a1n ho\u1eb7c b\u1eb1ng s\u1ed1 l\u1edbn nh\u1ea5t, GCD lu\u00f4n nh\u1ecf h\u01a1n ho\u1eb7c b\u1eb1ng s\u1ed1 nh\u1ecf nh\u1ea5t.<\/span><\/p>\n L\u1ea5y s\u1ed1 m\u0169 nh\u1ecf nh\u1ea5t thay v\u00ec l\u1edbn nh\u1ea5t khi ph\u00e2n t\u00edch th\u1eeba s\u1ed1 nguy\u00ean t\u1ed1. L\u1ea5y s\u1ed1 m\u0169 nh\u1ecf nh\u1ea5t c\u1ee7a c\u00e1c th\u1eeba s\u1ed1 chung l\u00e0 c\u00e1ch t\u00ecm GCD.<\/span><\/p>\n D\u00f9ng c\u00f4ng th\u1ee9c a \u00d7 b \u00f7 GCD cho ba s\u1ed1: c\u00f4ng th\u1ee9c ch\u1ec9 \u0111\u00fang v\u1edbi hai s\u1ed1. V\u1edbi ba s\u1ed1, t\u00ecm LCM c\u1ee7a hai s\u1ed1 tr\u01b0\u1edbc r\u1ed3i l\u1ea5y LCM c\u1ee7a k\u1ebft qu\u1ea3 v\u1edbi s\u1ed1 th\u1ee9 ba.<\/span><\/p>\n Vi\u1ec7c duy tr\u00ec th\u00f3i quen t\u1ef1 \u0111\u00e1nh gi\u00e1 sau m\u1ed7i b\u00e0i t\u1eadp s\u1ebd gi\u00fap h\u1ecdc sinh nh\u1eadn di\u1ec7n \u0111i\u1ec3m y\u1ebfu, kh\u1eafc ph\u1ee5c l\u1ed7i sai c\u00f3 h\u1ec7 th\u1ed1ng v\u00e0 n\u00e2ng cao \u0111\u1ed9 ch\u00ednh x\u00e1c khi l\u00e0m b\u00e0i.<\/span><\/p>\n C\u00e1ch luy\u1ec7n LCM hi\u1ec7u qu\u1ea3 l\u00e0 l\u00e0m theo ba t\u1ea7ng. \u0110\u1ea7u ti\u00ean, ch\u1ecdn c\u00e1c c\u1eb7p s\u1ed1 nh\u1ecf \u0111\u1ec3 ki\u1ec3m tra b\u1eb1ng c\u00e1ch li\u1ec7t k\u00ea b\u1ed9i. Sau \u0111\u00f3 chuy\u1ec3n sang ph\u00e2n t\u00edch th\u1eeba s\u1ed1 nguy\u00ean t\u1ed1 \u0111\u1ec3 nh\u00ecn ra v\u00ec sao ph\u1ea3i l\u1ea5y s\u1ed1 m\u0169 l\u1edbn nh\u1ea5t c\u1ee7a t\u1eebng th\u1eeba s\u1ed1. Cu\u1ed1i c\u00f9ng, l\u00e0m b\u00e0i to\u00e1n l\u1eddi v\u0103n v\u1ec1 chu k\u1ef3 ho\u1eb7c b\u00e0i li\u00ean quan \u0111\u1ebfn ph\u00e2n s\u1ed1. M\u1ed7i l\u1ea7n sai, ghi r\u00f5 m\u00ecnh \u0111\u00e3 nh\u1ea7m LCM v\u1edbi GCD, b\u1ecf s\u00f3t th\u1eeba s\u1ed1 hay di\u1ec5n gi\u1ea3i sai t\u00ecnh hu\u1ed1ng.<\/span><\/p>\n Gi\u1ea3 s\u1eed m\u1ed9t xe bu\u00fdt \u0111i qua tr\u1ea1m m\u1ed7i 6 ph\u00fat, xe kh\u00e1c \u0111i qua m\u1ed7i 8 ph\u00fat v\u00e0 c\u1ea3 hai v\u1eeba c\u00f9ng xu\u1ea5t ph\u00e1t. Th\u1eddi \u0111i\u1ec3m ch\u00fang g\u1eb7p l\u1ea1i \u1edf tr\u1ea1m l\u00e0 b\u1ed9i chung nh\u1ecf nh\u1ea5t c\u1ee7a 6 v\u00e0 8. C\u00e1c b\u1ed9i c\u1ee7a 6 l\u00e0 6, 12, 18, 24; c\u00e1c b\u1ed9i c\u1ee7a 8 l\u00e0 8, 16, 24. V\u1eady \u0111\u00e1p \u00e1n l\u00e0 24 ph\u00fat. N\u1ebfu ch\u1ec9 l\u1ea5y 6 nh\u00e2n 8, em \u0111\u01b0\u1ee3c 48 – v\u1eabn l\u00e0 m\u1ed9t th\u1eddi \u0111i\u1ec3m hai xe c\u00f9ng \u0111\u1ebfn, nh\u01b0ng kh\u00f4ng ph\u1ea3i l\u1ea7n \u0111\u1ea7u ti\u00ean. \u0110\u00e2y l\u00e0 l\u00fd do c\u1ea7n t\u00ecm b\u1ed9i chung nh\u1ecf nh\u1ea5t, kh\u00f4ng ch\u1ec9 m\u1ed9t b\u1ed9i chung b\u1ea5t k\u1ef3. Khi \u0111\u1ec1 cho chu k\u1ef3 c\u1ee7a ba s\u1ef1 ki\u1ec7n, \u00e1p d\u1ee5ng c\u00f9ng logic: t\u00ecm th\u1eddi \u0111i\u1ec3m nh\u1ecf nh\u1ea5t xu\u1ea5t hi\u1ec7n trong c\u1ea3 ba d\u00e3y b\u1ed9i s\u1ed1.<\/span><\/p>\n B\u00e0i t\u1eadp minh h\u1ecda k\u00e8m l\u1eddi gi\u1ea3i<\/b><\/p>\n \u0110\u1ec1 b\u00e0i: What is the least common multiple of 12 and 18?<\/span><\/p>\n L\u1eddi gi\u1ea3i:<\/span><\/p>\n Ph\u00e2n t\u00edch hai s\u1ed1 th\u00e0nh th\u1eeba s\u1ed1 nguy\u00ean t\u1ed1:<\/span><\/p>\n \u0110\u1ec3 t\u00ecm LCM, ta ch\u1ecdn l\u0169y th\u1eeba l\u1edbn nh\u1ea5t c\u1ee7a m\u1ed7i th\u1eeba s\u1ed1 nguy\u00ean t\u1ed1 xu\u1ea5t hi\u1ec7n trong hai s\u1ed1:\u00a0<\/span><\/p>\n LCM (12,18) = 2\u00b2 \u00d7 3\u00b2<\/span><\/p>\n \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= 4 x 9 = 36<\/span><\/p>\n \u0110\u00e1p \u00e1n \u0111\u00fang: B. 36\u00a0<\/span><\/p>\n \u0110\u1ec1 b\u00e0i: A student needs to solve the equation<\/span><\/p>\n x\/8 + x\/12 = 10<\/p>\n What is the least common multiple of the denominators 8 and 12?<\/span><\/p>\n B\u01b0\u1edbc 1: Ph\u00e2n t\u00edch c\u00e1c m\u1eabu s\u1ed1 th\u00e0nh th\u1eeba s\u1ed1 nguy\u00ean t\u1ed1:\u00a0<\/span><\/p>\n 8 =2\u00b3\u00a0 <\/span>, 12= 2\u00b2 x 3<\/span><\/p>\n B\u01b0\u1edbc 2: Ch\u1ecdn l\u0169y th\u1eeba l\u1edbn nh\u1ea5t c\u1ee7a t\u1eebng th\u1eeba s\u1ed1 nguy\u00ean t\u1ed1:\u00a0<\/span><\/p>\n LCM (8,12) = <\/span>2\u00b3 <\/span>x 3 = 8 x 3 = 24<\/span><\/p>\n V\u1eady b\u1ed9i chung nh\u1ecf nh\u1ea5t c\u1ee7a hai m\u1eabu s\u1ed1 l\u00e0 24. \u0110\u00e2y c\u0169ng l\u00e0 m\u1eabu s\u1ed1 chung nh\u1ecf nh\u1ea5t c\u00f3 th\u1ec3 d\u00f9ng \u0111\u1ec3 quy \u0111\u1ed3ng hai ph\u00e2n s\u1ed1 trong ph\u01b0\u01a1ng tr\u00ecnh tr\u00ean.<\/span><\/p>\n \u0110\u00e1p \u00e1n \u0111\u00fang: C. 24<\/span><\/p>\n \u0110\u1ec1 b\u00e0i: Two lights flash at regular intervals. The first light flashes every 6 seconds, and the second light flashes every 8 seconds. If both lights flash at the same time now, after how many seconds will they next flash at the same time?<\/span><\/p>\n L\u1eddi gi\u1ea3i:<\/span><\/p>\n B\u01b0\u1edbc 1: X\u00e1c \u0111\u1ecbnh d\u1ea1ng b\u00e0i.<\/span><\/p>\n Hai \u0111\u00e8n nh\u00e1y theo chu k\u1ef3 6 gi\u00e2y v\u00e0 8 gi\u00e2y. Mu\u1ed1n t\u00ecm th\u1eddi \u0111i\u1ec3m c\u1ea3 hai c\u00f9ng nh\u00e1y tr\u1edf l\u1ea1i, ta c\u1ea7n t\u00ecm b\u1ed9i chung nh\u1ecf nh\u1ea5t c\u1ee7a 6 v\u00e0 8.<\/span><\/p>\n B\u01b0\u1edbc 2: Ph\u00e2n t\u00edch th\u00e0nh th\u1eeba s\u1ed1 nguy\u00ean t\u1ed1:<\/span><\/p>\n 6=2 x <\/span>3, 8 = 2\u00b3<\/span><\/p>\n B\u01b0\u1edbc 3: T\u00ecm LCM:<\/span><\/p>\n LCM (6,8) = <\/span>2<\/span>3<\/span>x 3 = 8 x 3 = 24<\/span><\/p>\n V\u1eady sau 24 gi\u00e2y, hai \u0111\u00e8n s\u1ebd c\u00f9ng nh\u00e1y tr\u1edf l\u1ea1i.<\/span><\/p>\n \u0110\u00e1p \u00e1n \u0111\u00fang: B. 24 seconds<\/span><\/p>\n In<\/span> ETEST<\/span><\/a>, h\u1ecdc sinh \u0111\u01b0\u1ee3c c\u1ee7ng c\u1ed1 ki\u1ebfn th\u1ee9c n\u1ec1n t\u1ea3ng <\/span>SAT <\/span><\/a>Math nh\u01b0 Least Common Multiple (LCM) \u2013 b\u1ed9i chung nh\u1ecf nh\u1ea5t (BCNN) th\u00f4ng qua ph\u01b0\u01a1ng ph\u00e1p h\u1ecdc c\u00f3 h\u1ec7 th\u1ed1ng, b\u00e0i t\u1eadp theo t\u1eebng d\u1ea1ng v\u00e0 luy\u1ec7n \u0111\u1ec1 s\u00e1t v\u1edbi c\u1ea5u tr\u00fac <\/span>Digital SAT<\/span><\/a>. Kh\u00f4ng ch\u1ec9 ghi nh\u1edb c\u00f4ng th\u1ee9c, h\u1ecdc sinh c\u00f2n \u0111\u01b0\u1ee3c h\u01b0\u1edbng d\u1eabn c\u00e1ch nh\u1eadn di\u1ec7n LCM trong c\u00e1c b\u00e0i to\u00e1n ph\u00e2n s\u1ed1, ph\u01b0\u01a1ng tr\u00ecnh v\u00e0 chu k\u1ef3, t\u1eeb \u0111\u00f3 l\u1ef1a ch\u1ecdn ph\u01b0\u01a1ng ph\u00e1p gi\u1ea3i ph\u00f9 h\u1ee3p v\u00e0 t\u1ed1i \u01b0u th\u1eddi gian l\u00e0m b\u00e0i. V\u1edbi <\/span>l\u1ed9 tr\u00ecnh h\u1ecdc <\/span><\/a>\u0111\u01b0\u1ee3c x\u00e2y d\u1ef1ng theo n\u0103ng l\u1ef1c \u0111\u1ea7u v\u00e0o, ETEST \u0111\u1ed3ng h\u00e0nh c\u00f9ng h\u1ecdc sinh t\u1eebng b\u01b0\u1edbc n\u00e2ng cao t\u01b0 duy To\u00e1n h\u1ecdc, h\u1ea1n ch\u1ebf l\u1ed7i sai v\u00e0 h\u01b0\u1edbng \u0111\u1ebfn m\u1ee5c ti\u00eau \u0111i\u1ec3m<\/span> SAT<\/span><\/a> desire.\u00a0<\/span><\/p>\n LCM l\u00e0 b\u1ed9i chung nh\u1ecf nh\u1ea5t c\u1ee7a c\u00e1c s\u1ed1; LCD l\u00e0 m\u1eabu s\u1ed1 chung nh\u1ecf nh\u1ea5t c\u1ee7a c\u00e1c ph\u00e2n s\u1ed1. Khi quy \u0111\u1ed3ng ph\u00e2n s\u1ed1, LCD th\u01b0\u1eddng ch\u00ednh l\u00e0 LCM c\u1ee7a c\u00e1c m\u1eabu s\u1ed1.<\/span><\/p>\n N\u00ean nh\u1edb c\u00f4ng th\u1ee9c v\u1edbi hai s\u1ed1 nguy\u00ean d\u01b0\u01a1ng: GCD(a,b) \u00d7 LCM(a,b) = a \u00d7 b. Tuy nhi\u00ean, ch\u1ec9 d\u00f9ng khi \u0111\u1ec1 cho \u0111\u1ee7 d\u1eef ki\u1ec7n v\u00e0 em hi\u1ec3u m\u00ecnh \u0111ang t\u00ecm \u0111\u1ea1i l\u01b0\u1ee3ng n\u00e0o.<\/span><\/p>\n \u0110i\u1ec3m s\u1ed1 c\u1ee7a m\u1ed9t b\u00e0i luy\u1ec7n ch\u01b0a n\u00f3i h\u1ebft m\u1ee9c \u0111\u1ed9 ti\u1ebfn b\u1ed9. H\u00e3y theo d\u00f5i xem h\u1ecdc sinh c\u00f3 nh\u1eadn ra d\u1ea1ng b\u00e0i nhanh h\u01a1n, \u00edt l\u1eb7p l\u1ea1i c\u00f9ng m\u1ed9t l\u1ed7i h\u01a1n v\u00e0 c\u00f3 th\u1ec3 gi\u1ea3i th\u00edch c\u00e1ch l\u00e0m r\u00f5 r\u00e0ng h\u01a1n hay kh\u00f4ng. \u0110\u00f3 l\u00e0 nh\u1eefng d\u1ea5u hi\u1ec7u cho th\u1ea5y n\u1ec1n t\u1ea3ng \u0111ang v\u1eefng d\u1ea7n.<\/span><\/p>\n ETEST c\u00f3 th\u1ec3 gi\u00fap h\u1ecdc sinh c\u1ee7ng c\u1ed1 n\u1ec1n t\u1ea3ng Advanced Math theo d\u1ea1ng l\u1ed7i th\u1ef1c t\u1ebf, tr\u01b0\u1edbc khi chuy\u1ec3n sang c\u00e1c c\u00e2u c\u00f3 nhi\u1ec1u b\u01b0\u1edbc h\u01a1n.<\/span><\/p>\n\n\nLCM xu\u1ea5t hi\u1ec7n \u1edf \u0111\u00e2u trong Digital SAT?<\/b><\/h1>\n
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L\u1ed7i th\u01b0\u1eddng g\u1eb7p v\u1edbi LCM<\/b><\/h1>\n
C\u00e1ch t\u1ef1 ki\u1ec3m tra khi luy\u1ec7n<\/b><\/h1>\n
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C\u00e1ch luy\u1ec7n \u0111\u1ec3 ki\u1ebfn th\u1ee9c d\u00f9ng \u0111\u01b0\u1ee3c trong ph\u00f2ng thi<\/b><\/h1>\n
V\u00ed d\u1ee5 ng\u1eafn \u0111\u1ec3 hi\u1ec3u b\u1ea3n ch\u1ea5t c\u1ee7a LCM<\/b><\/h1>\n
B\u00e0i t\u1eadp 1: T\u00ecm b\u1ed9i chung nh\u1ecf nh\u1ea5t (C\u01a1 b\u1ea3n)<\/b><\/h2>\n
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B\u00e0i t\u1eadp 2: T\u00ecm LCM trong b\u00e0i to\u00e1n ph\u00e2n s\u1ed1 (V\u1eadn d\u1ee5ng)<\/b><\/h2>\n
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B\u00e0i t\u1eadp 3: B\u00e0i to\u00e1n chu k\u1ef3 (V\u1eadn d\u1ee5ng cao)<\/b><\/h2>\n
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C\u00e2u h\u1ecfi th\u01b0\u1eddng g\u1eb7p:<\/b><\/h1>\n
LCM v\u00e0 LCD kh\u00e1c nhau th\u1ebf n\u00e0o?<\/b><\/h2>\n
C\u00f3 c\u1ea7n h\u1ecdc thu\u1ed9c c\u00f4ng th\u1ee9c GCD nh\u00e2n LCM kh\u00f4ng?<\/b><\/h2>\n
\u0110i\u1ec1u h\u1ecdc sinh v\u00e0 ph\u1ee5 huynh n\u00ean theo d\u00f5i<\/b><\/h2>\n