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{"id":44335,"date":"2019-10-15T15:15:30","date_gmt":"2019-10-15T08:15:30","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=44335"},"modified":"2019-10-15T15:15:32","modified_gmt":"2019-10-15T08:15:32","slug":"giai-toan-10-chuong-4-bat-dang-thuc-bat-phuong-trinh-bai-2-bat-phuong-trinh-va-he-bat-phuong-trinh-mot-an","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/giai-toan-10-chuong-4-bat-dang-thuc-bat-phuong-trinh-bai-2-bat-phuong-trinh-va-he-bat-phuong-trinh-mot-an\/","title":{"rendered":"[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh\/ B\u00e0i 2: B\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh v\u00e0 h\u1ec7 b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh m\u1ed9t \u1ea9n"},"content":{"rendered":"\n

Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 2 trang 80<\/strong>: Cho m\u1ed9t v\u00ed d\u1ee5 v\u1ec1 b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh m\u1ed9t \u1ea9n, ch\u1ec9 r\u00f5 v\u1ebf tr\u00e1i v\u00e0 v\u1ebf ph\u1ea3i c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh n\u00e0y<\/ins><\/p>\n\n\n\n

L\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n

2x + 3 \u2265 -6<\/p>\n\n\n\n

V\u1ebf tr\u00e1i c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh: 2x + 3<\/p>\n\n\n\n

V\u1ebf ph\u1ea3i c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh: -6<\/p>\n\n\n\n

Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 2 trang 81<\/strong>: Cho b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh 2x \u2264 3.<\/ins><\/p>\n\n\n\n

a) Trong c\u00e1c s\u1ed1 -2; 2 1\/2; \u03c0; \u221a10 s\u1ed1 n\u00e0o l\u00e0 nghi\u1ec7m, s\u1ed1 n\u00e0o kh\u00f4ng l\u00e0 nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh tr\u00ean ?<\/p>\n\n\n\n

b) Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00f3 v\u00e0 bi\u1ec3u di\u1ec5n t\u1eadp nghi\u1ec7m c\u1ee7a n\u00f3 tr\u00ean tr\u1ee5c s\u1ed1.<\/p>\n\n\n\n

L\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n

a) C\u00e1c s\u1ed1 l\u00e0 nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh tr\u00ean l\u00e0: -2; <\/p>\n\n\n\n

C\u00e1c s\u1ed1 kh\u00f4ng l\u00e0 nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh tr\u00ean l\u00e0: 2 1\/2; \u03c0; \u221a10<\/p>\n\n\n\n

b)2x \u2264 3 \u21d4 x \u2264 3\/2<\/p>\n\n\n\n

Bi\u1ec3u di\u1ec5n t\u1eadp nghi\u1ec7m tr\u00ean tr\u1ee5c s\u1ed1 l\u00e0:<\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 2 trang 82<\/strong>: Hai b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh trong v\u00ed d\u1ee5 1 c\u00f3 t\u01b0\u01a1ng \u0111\u01b0\u01a1ng hay kh\u00f4ng ? V\u00ec sao ?<\/ins><\/p>\n\n\n\n

L\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n

Hai b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh trong VD 1 kh\u00f4ng t\u01b0\u01a1ng \u0111\u01b0\u01a1ng do ch\u00fang kh\u00f4ng c\u00f3 c\u00f9ng t\u1eadp nghi\u1ec7m.<\/p>\n\n\n\n

B\u00e0i 1 (trang 87 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: T\u00ecm c\u00e1c gi\u00e1 tr\u1ecb x th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n c\u1ee7a m\u1ed7i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh sau:<\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

L\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

V\u1eady t\u1eadp gi\u00e1 tr\u1ecb c\u1ee7a x th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh l\u00e0 D = R\\{0; \u20131}<\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

BPT x\u00e1c \u0111\u1ecbnh khi <\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

V\u1eady t\u1eadp gi\u00e1 tr\u1ecb c\u1ee7a x th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh l\u00e0 D = R\\{\u20132; 1; 2; 3}<\/ins><\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

BPT x\u00e1c \u0111\u1ecbnh khi x + 1 \u2260 0 \u21d4 x \u2260 \u20131. <\/p>\n\n\n\n

V\u1eady t\u1eadp gi\u00e1 tr\u1ecb c\u1ee7a x th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh l\u00e0 D = R\\{\u20131}<\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

V\u1eady t\u1eadp gi\u00e1 tr\u1ecb c\u1ee7a x th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh l\u00e0 D = (\u2013\u221e; 1] \\ {\u20134}. <\/p>\n\n\n\n

Ki\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n

T\u00ecm t\u1eadp x\u00e1c \u0111\u1ecbnh c\u1ee7a BPT c\u00f9ng gi\u1ed1ng t\u00ecm t\u1eadp x\u00e1c \u0111\u1ecbnh c\u1ee7a PT<\/p>\n\n\n\n

BPT x\u00e1c \u0111\u1ecbnh khi m\u1ed7i bi\u1ec3u th\u1ee9c trong n\u00f3 x\u00e1c \u0111\u1ecbnh. <\/p>\n\n\n\n

Bi\u1ec3u th\u1ee9c ch\u1ee9a c\u0103n b\u1eadc ch\u1eb5n x\u00e1c \u0111\u1ecbnh khi bi\u1ec3u th\u1ee9c trong c\u0103n \u2265 0 <\/p>\n\n\n\n

Ph\u00e2n th\u1ee9c x\u00e1c \u0111\u1ecbnh khi bi\u1ec3u th\u1ee9c \u1edf m\u1eabu th\u1ee9c kh\u00e1c 0.<\/p>\n\n\n\n

B\u00e0i 2 (trang 88 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Ch\u1ee9ng minh c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh sau v\u00f4 nghi\u1ec7m:<\/ins><\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

L\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n

a) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh x \u2265 \u20138<\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n
\"Gi\u1ea3i<\/figure>\n\n\n\n

Ta c\u00f3: \n n\u00ean \n v\u1edbi m\u1ecdi x \u2265 \u20138. <\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

Do \u0111\u00f3 BPT \n v\u00f4 nghi\u1ec7m. <\/p>\n\n\n\n

b) T\u1eadp x\u00e1c \u0111\u1ecbnh: D = R. <\/ins><\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n
\"Gi\u1ea3i<\/figure>\n\n\n\n

Do \u0111\u00f3 BPT \n v\u00f4 nghi\u1ec7m. <\/p>\n\n\n\n

c) T\u1eadp x\u00e1c \u0111\u1ecbnh D = R. <\/p>\n\n\n\n

Ta c\u00f3:<\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

B\u00e0i 3 (trang 88 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i th\u00edch v\u00ec sao c\u00e1c c\u1eb7p b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh sau t\u01b0\u01a1ng \u0111\u01b0\u01a1ng?<\/p>\n\n\n\n

a) -4x + 1 > 0 v\u00e0 4x – 1 < 0<\/ins><\/p>\n\n\n\n

b) 2x2<\/sup> + 5 \u2264 2x – 1 v\u00e0 2x2<\/sup> – 2x + 6 \u2264 0<\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

L\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n

a) Nh\u00e2n hai v\u1ebf c\u1ee7a BPT \u20134x + 1 > 0 v\u1edbi (\u20131) ta \u0111\u01b0\u1ee3c BPT 4x \u2013 1 < 0 n\u00ean hai BPT \u0111\u00f3 t\u01b0\u01a1ng \u0111\u01b0\u01a1ng. <\/p>\n\n\n\n

Vi\u1ebft l\u00e0 \u20134x + 1 > 0 \u21d4 4x \u2013 1 < 0.<\/p>\n\n\n\n

b) Ta c\u00f3: <\/p>\n\n\n\n

2x2<\/sup> + 5 \u2264 2x \u2013 1<\/p>\n\n\n\n

\u21d4 2x2<\/sup> + 5 + 1 \u2013 2x \u2264 2x \u2013 1 + 1 \u2013 2x (C\u1ed9ng c\u1ea3 hai v\u1ebf c\u1ee7a BPT v\u1edbi 1 \u2013 2x). <\/p>\n\n\n\n

\u21d4 2x2<\/sup> \u2013 2x + 6 \u2264 0.<\/p>\n\n\n\n

V\u1eady hai BPT 2x2<\/sup> + 5 \u2264 2x \u2013 1 \u21d4 2x2<\/sup> \u2013 2x + 6 \u2264 0. <\/ins><\/p>\n\n\n\n

c) Ta c\u00f3: x + 1 > 0<\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

d) \u0110i\u1ec1u ki\u1ec7n x \u2265 1, khi \u0111\u00f3 2x + 1 > 0.<\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

Ki\u1ebfn th\u1ee9c \u00e1p d\u1ee5ng<\/strong><\/p>\n\n\n\n

Khi s\u1eed d\u1ee5ng c\u00e1c ph\u00e9p bi\u1ebfn \u0111\u1ed5i t\u01b0\u01a1ng \u0111\u01b0\u01a1ng ta nh\u1eadn \u0111\u01b0\u1ee3c c\u00e1c BPT t\u01b0\u01a1ng \u0111\u01b0\u01a1ng. <\/p>\n\n\n\n

C\u00e1c ph\u00e9p bi\u1ebfn \u0111\u1ed5i t\u01b0\u01a1ng \u0111\u01b0\u01a1ng g\u1ed3m: <\/p>\n\n\n\n

+ C\u1ed9ng ho\u1eb7c tr\u1eeb hai v\u1ebf c\u1ee7a BPT v\u1edbi c\u00f9ng m\u1ed9t bi\u1ec3u th\u1ee9c: <\/p>\n\n\n\n

     P(x) < Q(x) \u21d4 P(x) + f(x) < Q(x) + f(x).<\/p>\n\n\n\n

+ Nh\u00e2n ho\u1eb7c chia hai v\u1ebf c\u1ee7a BPT v\u1edbi c\u00f9ng m\u1ed9t bi\u1ec3u th\u1ee9c kh\u00e1c 0. <\/p>\n\n\n\n

     P (x) < Q(x) \u21d4 P(x).f(x) < Q(x).f(x) n\u1ebfu f(x) > 0<\/p>\n\n\n\n

     P(x) < Q(x) \u21d4 P(x).f(x) > Q(x).f(x) n\u1ebfu f(x) < 0.<\/p>\n\n\n\n

+ N\u00e2ng l\u00ean l\u0169y th\u1eeba b\u1eadc ch\u1eb5n c\u1ee7a BPT c\u00f3 c\u1ea3 hai v\u1ebf \u0111\u1ec1u d\u01b0\u01a1ng: <\/p>\n\n\n\n

     0 < P(x) < Q(x) \u21d4 P2n<\/sup>(x) < Q2n<\/sup>(x)<\/p>\n\n\n\n

+ N\u00e2ng l\u00ean l\u0169y th\u1eeba b\u1eadc l\u1ebb c\u1ea3 hai v\u1ebf c\u1ee7a BPT <\/p>\n\n\n\n

     P(x) < Q(x) \u21d4 P2n+1<\/sup>(x) < Q2n+1<\/sup>(x).<\/p>\n\n\n\n

+ Khai c\u0103n b\u1eadc hai c\u1ee7a BPT c\u00f3 c\u1ea3 hai v\u1ebf \u0111\u1ec1u d\u01b0\u01a1ng : <\/p>\n\n\n\n

     0 < P(x) < Q(x) \u21d4 \u221aP(x) < \u221aQ(x)<\/p>\n\n\n\n

+ Khai c\u0103n b\u1eadc ba c\u1ea3 hai v\u1ebf c\u1ee7a BPT : <\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

B\u00e0i 4 (trang 88 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i c\u00e1c b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh sau:<\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

<\/ins><\/p>\n\n\n\n

b. (2x – 1)(x + 3) – 3x + 1 \u2264 (x – 1)(x + 3) + x2<\/sup> – 5<\/p>\n\n\n\n

L\u1eddi gi\u1ea3i<\/strong><\/p>\n\n\n\n

a) T\u1eadp x\u00e1c \u0111\u1ecbnh D = R. <\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n
\"Gi\u1ea3i<\/figure>\n\n\n\n

V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \n<\/p>\n\n\n\n

b) (2x \u2013 1)(x + 3) \u2013 3x + 1 \u2264 (x \u2013 1)(x + 3) + x2 \u2013 5<\/p>\n\n\n\n

\u21d4 2×2 \u2013 x + 6x \u2013 3 \u2013 3x + 1 \u2264 x2 \u2013 x + 3x \u2013 3 + x2 \u2013 5 <\/p>\n\n\n\n

\u21d4 2×2 + 2x \u2013 2 \u2264 2×2 + 2x \u2013 8 <\/p>\n\n\n\n

\u21d4 6 \u2264 0 (V\u00f4 l\u00fd). <\/p>\n\n\n\n

V\u1eady BPT v\u00f4 nghi\u1ec7m. <\/p>\n\n\n\n

B\u00e0i 5 (trang 88 SGK \u0110\u1ea1i S\u1ed1 10)<\/strong>: Gi\u1ea3i h\u1ec7 b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh sau:<\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

L\u1eddi gi\u1ea3i<\/strong><\/ins><\/p>\n\n\n\n

a) T\u1eadp x\u00e1c \u0111\u1ecbnh D = R. <\/p>\n\n\n\n

Gi\u1ea3i t\u1eebng b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh ta c\u00f3: <\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

<\/ins><\/p>\n\n\n\n

V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a h\u1ec7 b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0<\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

b) T\u1eadp x\u00e1c \u0111\u1ecbnh D = R. <\/p>\n\n\n\n

Gi\u1ea3i t\u1eebng b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh: <\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n\n\n\n

V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a h\u1ec7 b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 <\/p>\n\n\n\n

\"Gi\u1ea3i<\/figure>\n","protected":false},"excerpt":{"rendered":"

Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 2 trang 80: Cho m\u1ed9t v\u00ed d\u1ee5 v\u1ec1 b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh m\u1ed9t \u1ea9n, ch\u1ec9 r\u00f5 v\u1ebf tr\u00e1i v\u00e0 v\u1ebf ph\u1ea3i c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh n\u00e0y L\u1eddi gi\u1ea3i 2x + 3 \u2265 -6 V\u1ebf tr\u00e1i c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh: 2x + 3 V\u1ebf ph\u1ea3i c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng […]<\/p>\n","protected":false},"author":12,"featured_media":44336,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1633],"tags":[],"yoast_head":"\n[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 4: B\u1ea5t \u0111\u1eb3ng th\u1ee9c. 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