Ghi ch\u00fa: T\u00e0i li\u1ec7u tr\u00edch d\u1eabn<\/p>\n\n\n\n
\u0110a th\u1ee9c m\u1ed9t bi\u1ebfn l\u00e0 m\u1ed9t trong nh\u1eefng ki\u1ebfn th\u1ee9c n\u1ec1n t\u1ea3ng to\u00e1n h\u1ecdc quan tr\u1ecdng. \u0110\u1ec3 hi\u1ec3u \u0111\u01b0\u1ee3c l\u00fd thuy\u1ebft, c\u00e1c d\u1ea1ng b\u00e0i t\u1eadp v\u1ec1 c\u1ed9ng tr\u1eeb \u0111a th\u1ee9c m\u1ed9t bi\u1ebfn l\u1edbp 7 c\u0169ng nh\u01b0 c\u00e1ch s\u1eafp x\u1ebfp theo l\u0169y th\u1eeba t\u0103ng ho\u1eb7c gi\u1ea3m d\u1ea7n c\u1ee7a bi\u1ebfn. H\u00e3y c\u00f9ng Vietlearn.org\/ theo d\u00f5i b\u00e0i gi\u1ea3ng sau.<\/p>\n\n\n\n
N\u1ed9i dung b\u00e0i gi\u1ea3ng \u0111a th\u1ee9c m\u1ed9t bi\u1ebfn<\/p>\n\n\n\n
I. Kh\u00e1i ni\u1ec7m \u0111a th\u1ee9c m\u1ed9t bi\u1ebfn<\/p>\n\n\n\n
\u0110a th\u1ee9c m\u1ed9t bi\u1ebfn ch\u00ednh l\u00e0 t\u1ed5ng c\u1ee7a c\u00e1c \u0111\u01a1n th\u1ee9c trong c\u00f9ng m\u1ed9t bi\u1ebfn.<\/p>\n\n\n\n
Tuy nhi\u00ean c\u1ea7n l\u01b0u \u00fd r\u1eb1ng trong to\u00e1n h\u1ecdc, m\u1ed7i s\u1ed1 s\u1ebd \u0111\u01b0\u1ee3c coi nh\u01b0 \u0111a th\u1ee9c 1 bi\u1ebfn.<\/p>\n\n\n\n
II. B\u1eadc c\u1ee7a \u0111a th\u1ee9c 1 bi\u1ebfn<\/p>\n\n\n\n
B\u1eadc c\u1ee7a \u0111a th\u1ee9c 1 bi\u1ebfn l\u00e0 s\u1ed1 m\u0169 c\u00f3 s\u1ed1 h\u1ea1ng nh\u1ea5t c\u1ee7a bi\u1ebfn thu\u1ed9c \u0111a th\u1ee9c \u0111\u00f3. (Kh\u00e1c v\u1edbi \u0111a th\u1ee9c 0 \u0111\u00e3 th\u1ef1c hi\u1ec7n thu g\u1ecdn).<\/p>\n\n\n\n
III. H\u1ec7 s\u1ed1 v\u00e0 gi\u00e1 tr\u1ecb c\u1ee7a 1 \u0111a th\u1ee9c<\/p>\n\n\n\n
H\u1ec7 s\u1ed1 c\u1ee7a m\u1ed9t \u0111a th\u1ee9c (\u0111\u00e3 th\u1ef1c hi\u1ec7n thu g\u1ecdn) ch\u00ednh l\u00e0 h\u1ec7 s\u1ed1 c\u1ee7a s\u1ed1 h\u1ea1ng m\u00e0 c\u00f3 b\u1eadc cao nh\u1ea5t. Th\u00f4ng th\u01b0\u1eddng h\u1ec7 s\u1ed1 t\u1ef1 do s\u1ebd l\u00e0 s\u1ed1 h\u1ea1ng m\u00e0 kh\u00f4ng ch\u1ee9a bi\u1ebfn.<\/p>\n\n\n\n
V\u00ed d\u1ee5 minh h\u1ecda:<\/p>\n\n\n\n
Cho \u0111a th\u1ee9c sau: f(x)= 5×4 + 12×3 \u2013 33x + 22<\/p>\n\n\n\n
H\u1ec7 s\u1ed1 c\u1ee7a \u0111a th\u1ee9c ch\u00ednh l\u00e0 5 v\u00e0 h\u1ec7 s\u1ed1 t\u1ef1 do ch\u00ednh l\u00e0 22<\/p>\n\n\n\n
Gi\u00e1 tr\u1ecb c\u1ee7a 1 \u0111a th\u1ee9c f(x) v\u1edbi x = a s\u1ebd \u0111\u01b0\u1ee3c k\u00ed hi\u1ec7u l\u00e0 f(a). \u0110\u1ec3 t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a 1 \u0111a th\u1ee9c c\u1ea7n thay x = a v\u00e0o trong \u0111a th\u1ee9c f(x) \u0111\u00e3 cho, sau \u0111\u00f3 thu g\u1ecdn l\u1ea1i.<\/p>\n\n\n\n
IV. Ph\u01b0\u01a1ng ph\u00e1p c\u1ed9ng, tr\u1eeb \u0111a th\u1ee9c 1 bi\u1ebfn:<\/p>\n\n\n\n
\u0110\u1ec3 th\u1ef1c hi\u1ec7n c\u1ed9ng tr\u1eeb \u0111a th\u1ee9c 1 bi\u1ebfn s\u1ebd c\u00f3 2 ph\u01b0\u01a1ng ph\u00e1p th\u1ee9c \u0111\u01a1n gi\u1ea3n sau:<\/p>\n\n\n\n
Ph\u01b0\u01a1ng ph\u00e1p 1: C\u1ed9ng tr\u1eeb \u0111a th\u1ee9c 1 bi\u1ebfn theo h\u00e0ng ngang.<\/p>\n\n\n\n
Ph\u01b0\u01a1ng ph\u00e1p 2: Th\u1ef1c hi\u1ec7n s\u1eafp x\u1ebfp ch\u00fang theo c\u00f9ng m\u1ed9t l\u0169y th\u1eeba gi\u1ea3m d\u1ea7n ho\u1eb7c t\u0103ng d\u1ea7n c\u1ee7a bi\u1ebfn. Sau \u0111\u00f3 \u0111\u1eb7t ph\u00e9p t\u00ednh v\u1edbi c\u1ed9t d\u1ecdc t\u01b0\u01a1ng \u1ee9ng, sau \u0111\u00f3 th\u1ef1c hi\u1ec7n t\u01b0\u01a1ng t\u1ef1 nh\u01b0 c\u1ed9ng tr\u1eeb c\u00e1c s\u1ed1. Tuy nhi\u00ean v\u1edbi ph\u01b0\u01a1ng ph\u00e1p n\u00e0y c\u1ea7n l\u01b0u \u00fd r\u1eb1ng ph\u1ea3i \u0111\u1eb7t nh\u1eefng \u0111\u01a1n th\u1ee9c c\u00f3 c\u00f9ng m\u1ed9t d\u1ea1ng \u1edf c\u00f9ng m\u1ed9t c\u1ed9t v\u1edbi nhau.<\/p>\n\n\n\n
Ph\u01b0\u01a1ng ph\u00e1p c\u1ed9ng, tr\u1eeb \u0111a th\u1ee9c 1 bi\u1ebfn<\/p>\n\n\n\n
V\u00ed d\u1ee5 minh h\u1ecda: Cho 2 \u0111a th\u1ee9c sau P(x) = 5×4 + 12×3 \u2013 33x + 22; Q(x) = 6 \u2013 2x + 5×3 \u2013 2×4. T\u00ednh P(x) \u2013 Q(x).<\/p>\n\n\n\n
H\u01b0\u1edbng d\u1eabn gi\u1ea3i chi ti\u1ebft:<\/p>\n\n\n\n
P(x) \u2013 Q(x) = (5×4 + 12×3 \u2013 33x + 22) \u2013 (6 \u2013 2x + 5×3 \u2013 2×4)<\/p>\n\n\n\n
= 5×4 + 12×3 \u2013 33x + 22 \u2013 6 + 2x \u2013 5×3 + 2×4<\/p>\n\n\n\n
= (5×4 + 2×4) + (12×3 \u2013 5×3)+ (- 33x + 2x) + (22 \u2013 6)<\/p>\n\n\n\n
= 7×4 + 7×3 \u2013 31x + 16<\/p>\n\n\n\n
Xem th\u00eam : L\u00fd thuy\u1ebft v\u00e0 l\u1eddi gi\u1ea3i b\u00e0i nghi\u1ec7m c\u1ee7a \u0111a th\u1ee9c 1 bi\u1ebfn<\/p><\/blockquote>\n\n\n\n
V. C\u00e1c d\u1ea1ng b\u00e0i t\u1eadp v\u1ec1 \u0111a th\u1ee9c 1 bi\u1ebfn:<\/p>\n\n\n\n
\u0110a th\u1ee9c 1 bi\u1ebfn l\u00e0 ki\u1ebfn th\u1ee9c n\u1ec1n t\u1ea3ng quan tr\u1ecdng trong ch\u01b0\u01a1ng tr\u00ecnh To\u00e1n<\/a> h\u1ecdc l\u1edbp 7. Ch\u00ednh v\u00ec v\u1eady tr\u01b0\u1edbc khi t\u00ecm hi\u1ec3u c\u00e1c d\u1ea1ng b\u00e0i t\u1eadp l\u00fd thuy\u1ebft v\u00e0 c\u1ed9ng tr\u1eeb \u0111a th\u1ee9c 1 bi\u1ebfn h\u00e3y c\u00f9ng h\u1ec7 th\u1ed1ng h\u00f3a l\u1ea1i ph\u01b0\u01a1ng ph\u00e1p qua s\u01a1 \u0111\u1ed3 t\u01b0 duy sau.<\/p>\n\n\n\n
S\u01a1 \u0111\u1ed3 t\u01b0 duy v\u1ec1 c\u1ed9ng tr\u1eeb \u0111a th\u1ee9c 1 bi\u1ebfn<\/p>\n\n\n\n
- B\u00e0i t\u1eadp l\u00fd thuy\u1ebft \u0111a th\u1ee9c 1 bi\u1ebfn:<\/li><\/ol>\n\n\n\n
B\u00e0i t\u1eadp l\u00fd thuy\u1ebft \u0111a th\u1ee9c m\u1ed9t bi\u1ebfn s\u1ebd gi\u00fap c\u00e1c b\u1ea1n h\u1ecdc sinh<\/a> n\u1eafm v\u1eefng \u0111\u01b0\u1ee3c k\u00fd hi\u1ec7u c\u1ee7a \u0111a th\u1ee9c 1 bi\u1ebfn c\u0169ng nh\u01b0 v\u1eadn d\u1ee5ng \u0111\u01b0\u1ee3c \u0111\u1ec3 gi\u1ea3i b\u00e0i t\u1eadp trong s\u00e1ch gi\u00e1o khoa.<\/p>\n\n\n\n
B\u00e0i t\u1eadp 1: (B\u00e0i 39\/SGK \u0110\u1ea1i s\u1ed1 l\u1edbp 7, trang 43, t\u1eadp 2)<\/p>\n\n\n\n
Cho \u0111a th\u1ee9c d\u01b0\u1edbi d\u1ea1ng m\u1ed9t bi\u1ebfn sau: P(x) = 2 + 5×2 \u2013 3×3 + 4×2 \u2013 2x \u2013 x3 + 6×5<\/p>\n\n\n\n
Thu g\u1ecdn, s\u1eafp x\u1ebfp h\u1ea1ng t\u1eed c\u1ee7a P(x) \u0111\u00e3 cho tr\u00ean v\u1edbi l\u0169y th\u1eeba gi\u1ea3m c\u1ee7a bi\u1ebfn.<\/p>\n\n\n\n
Vi\u1ebft h\u1ec7 s\u1ed1 kh\u00e1c 0 trong \u0111a th\u1ee9c P(x) \u0111\u00e3 cho tr\u00ean,<\/p>\n\n\n\n
B\u00e0i t\u1eadp 2: (B\u00e0i 40\/SGK \u0110\u1ea1i s\u1ed1 l\u1edbp 7, trang 43, t\u1eadp 2)<\/p>\n\n\n\n
Cho \u0111a th\u1ee9c d\u01b0\u1edbi d\u1ea1ng m\u1ed9t bi\u1ebfn sau: Q(x) = x2 + 2×4 + 4×3 \u2013 5×6 + 3×2 \u2013 4x \u2013 1<\/p>\n\n\n\n
S\u1eafp x\u1ebfp h\u1ea1ng t\u1eed c\u1ee7a Q(x) \u0111\u00e3 cho tr\u00ean v\u1edbi l\u0169y th\u1eeba gi\u1ea3m c\u1ee7a bi\u1ebfn.<\/p>\n\n\n\n
Ch\u1ec9 ra nh\u1eefng h\u1ec7 s\u1ed1 kh\u00e1c 0 trong \u0111a th\u1ee9c Q(x) \u0111\u00e3 cho tr\u00ean.<\/p>\n\n\n\n
B\u00e0i t\u1eadp 3: (B\u00e0i 41\/SGK \u0110\u1ea1i s\u1ed1 l\u1edbp 7, trang 43, t\u1eadp 2)<\/p>\n\n\n\n
Vi\u1ebft ra 1 \u0111a th\u1ee9c 1 bi\u1ebfn c\u00f3 2 h\u1ea1ng t\u1eed v\u00e0 5 l\u00e0 h\u1ec7 s\u1ed1 cao nh\u1ea5t, -1 l\u00e0 h\u1ec7 s\u1ed1 t\u1ef1 do c\u1ee7a \u0111a th\u1ee9c \u0111\u00f3.<\/p>\n\n\n\n
- B\u00e0i t\u1eadp v\u1ec1 c\u1ed9ng tr\u1eeb \u0111a th\u1ee9c 1 bi\u1ebfn:<\/li><\/ol>\n\n\n\n
B\u00e0i t\u1eadp v\u1ec1 c\u1ed9ng, tr\u1eeb \u0111a th\u1ee9c m\u1ed9t bi\u1ebfn s\u1ebd gi\u00fap c\u00e1c b\u1ea1n h\u1ecdc sinh<\/a> n\u1eafm v\u1eefng \u0111\u01b0\u1ee3c ki\u1ebfn th\u1ee9c ch\u01b0\u01a1ng c\u0169ng nh\u01b0 v\u1eadn d\u1ee5ng \u0111\u01b0\u1ee3c c\u00e1c ph\u01b0\u01a1ng ph\u00e1p \u0111\u1ec3 th\u1ef1c h\u00e0nh v\u00e0 \u00f4n luy\u1ec7n hi\u1ec7u qu\u1ea3.<\/p>\n\n\n\n
B\u00e0i t\u1eadp v\u1ec1 c\u1ed9ng, tr\u1eeb \u0111a th\u1ee9c 1 bi\u1ebfn<\/p>\n\n\n\n
B\u00e0i t\u1eadp 1: (B\u00e0i 44\/SGK \u0110\u1ea1i s\u1ed1 l\u1edbp 7, trang 45, t\u1eadp 2)<\/p>\n\n\n\n
Cho hai \u0111a th\u1ee9c d\u01b0\u1edbi d\u1ea1ng m\u1ed9t bi\u1ebfn sau:<\/p>\n\n\n\n
P(x) = -5×3 \u2013 \u2153 + 8×4 + x2<\/p>\n\n\n\n
Q(x) = x2 \u2013 5x \u2013 2×3 + x4 \u2013 \u2154<\/p>\n\n\n\n
H\u00e3y t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a P(x) \u2013 Q(x) v\u00e0 P(x) + Q(x) d\u1ef1a v\u00e0o 2 \u0111a th\u1ee9c \u0111\u00e3 cho tr\u00ean.<\/p>\n\n\n\n
B\u00e0i t\u1eadp 2: (B\u00e0i 45\/SGK \u0110\u1ea1i s\u1ed1 l\u1edbp 7, trang 45, t\u1eadp 2)<\/p>\n\n\n\n
Cho \u0111a th\u1ee9c d\u01b0\u1edbi d\u1ea1ng m\u1ed9t bi\u1ebfn sau: P(x) = 4×4 \u2013 3×2 + \u00bd \u2013 x<\/p>\n\n\n\n
T\u00ecm \u0111a th\u1ee9c R(x) v\u00e0 Q(x) d\u1ef1a v\u00e0o 2 \u0111a th\u1ee9c \u0111\u00e3 cho tr\u00ean sao cho:<\/p>\n\n\n\n
Q(x) + P(x) = x5 \u2013 2×2 + 1<\/p>\n\n\n\n
P(x) \u2013 R(x) = x3<\/p>\n\n\n\n
B\u00e0i t\u1eadp 3: (B\u00e0i 46\/SGK \u0110\u1ea1i s\u1ed1 l\u1edbp 7, trang 45, t\u1eadp 2)<\/p>\n\n\n\n
Cho \u0111a th\u1ee9c sau P(x) = 5×3 \u2013 4×2 + 7x \u2013 2, vi\u1ebft \u0111a th\u1ee9c \u0111\u00e3 cho sau d\u01b0\u1edbi d\u1ea1ng<\/p>\n\n\n\n
T\u1ed5ng c\u1ee7a 2 \u0111a th\u1ee9c 1 bi\u1ebfn \u0111\u00e3 cho<\/p>\n\n\n\n
Hi\u1ec7u c\u1ee7a 2 \u0111a th\u1ee9c 1 bi\u1ebfn \u0111\u00e3 cho<\/p>\n\n\n\n
B\u1ea1n Vinh \u0111\u01b0a ra nh\u1eadn x\u00e9t sau: \u201cC\u00f3 th\u1ec3 vi\u1ebft \u0111a th\u1ee9c P(x) th\u00e0nh t\u1ed5ng c\u1ee7a 2 \u0111a th\u1ee9c b\u1eadc 4\u201d l\u00e0 \u0111\u00fang hay sai v\u00e0 v\u00ec sao?<\/p>\n\n\n\n
\u0110a th\u1ee9c m\u1ed9t bi\u1ebfn v\u00e0 c\u00e1c d\u1ea1ng to\u00e1n th\u01b0\u1eddng g\u1eb7p sau l\u00e0 ki\u1ebfn th\u1ee9c n\u1ec1n t\u1ea3ng v\u00f4 c\u00f9ng quan tr\u1ecdng trong ch\u01b0\u01a1ng tr\u00ecnh to\u00e1n<\/a> h\u1ecdc l\u1edbp 7 v\u00e0 ch\u01b0\u01a1ng tr\u00ecnh h\u1ecdc v\u1ec1 sau. Hy v\u1ecdng v\u1edbi nh\u1eefng ki\u1ebfn th\u1ee9c \u0111\u01b0\u1ee3c cung c\u1ea5p tr\u00ean s\u1ebd l\u00e0 ngu\u1ed3n t\u00e0i li\u1ec7u b\u1ed5 \u00edch gi\u00fap c\u00e1c em h\u1ecdc sinh<\/a> v\u00e0 c\u00e1c b\u1eadc ph\u1ee5 huynh \u00f4n t\u1eadp \u0111\u01b0\u1ee3c hi\u1ec7u qu\u1ea3. Ngo\u00e0i ra n\u1ebfu c\u00f3 nhu c\u1ea7u t\u00ecm ki\u1ebfm th\u00eam th\u00f4ng tin c\u1ee7a m\u00f4n h\u1ecdc, h\u00e3y th\u01b0\u1eddng xuy\u00ean truy c\u1eadp Vietlearn.org\/ \u0111\u1ec3 n\u1eb1m b\u1eaft \u0111\u01b0\u1ee3c nh\u1eefng b\u00e0i gi\u1ea3ng m\u1edbi nh\u1ea5t nh\u00e9!<\/p>\n\n\n\n
<\/p>\n\n\n\n
H\u01b0\u1edbng d\u1eabn gi\u1ea3i b\u00e0i t\u1eadp ph\u00e9p tr\u1eeb hai s\u1ed1 nguy\u00ean l\u1edbp 6<\/p>\n\n\n\n
Ph\u00e9p tr\u1eeb c\u00e1c ph\u00e2n th\u1ee9c \u0111\u1ea1i s\u1ed1 \u2013 \u00d4n luy\u1ec7n to\u00e1n h\u1ecdc c\u00f9ng Topp<\/p>\n\n\n\n\n\n\n","protected":false},"excerpt":{"rendered":"
Ghi ch\u00fa: T\u00e0i li\u1ec7u tr\u00edch d\u1eabn \u0110a th\u1ee9c m\u1ed9t bi\u1ebfn l\u00e0 m\u1ed9t trong nh\u1eefng ki\u1ebfn th\u1ee9c n\u1ec1n t\u1ea3ng to\u00e1n h\u1ecdc quan tr\u1ecdng. \u0110\u1ec3 hi\u1ec3u \u0111\u01b0\u1ee3c l\u00fd thuy\u1ebft, c\u00e1c d\u1ea1ng b\u00e0i t\u1eadp v\u1ec1 c\u1ed9ng tr\u1eeb \u0111a th\u1ee9c m\u1ed9t bi\u1ebfn l\u1edbp 7 c\u0169ng nh\u01b0 c\u00e1ch s\u1eafp x\u1ebfp theo l\u0169y th\u1eeba t\u0103ng ho\u1eb7c gi\u1ea3m d\u1ea7n c\u1ee7a bi\u1ebfn. H\u00e3y […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":"","_disable_autopaging":false},"categories":[9],"tags":[],"class_list":["post-1029","post","type-post","status-publish","format-standard","hentry","category-toan-hoc"],"_links":{"self":[{"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/posts\/1029","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/comments?post=1029"}],"version-history":[{"count":0,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/posts\/1029\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/media?parent=1029"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/categories?post=1029"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/tags?post=1029"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}