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{"id":44320,"date":"2019-10-15T14:14:16","date_gmt":"2019-10-15T07:14:16","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=44320"},"modified":"2019-10-15T14:14:18","modified_gmt":"2019-10-15T07:14:18","slug":"giai-toan-10-chuong-2-ham-so-bac-nhat-va-bac-hai-bai-3-ham-so-bac-hai","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/giai-toan-10-chuong-2-ham-so-bac-nhat-va-bac-hai-bai-3-ham-so-bac-hai\/","title":{"rendered":"[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 2: H\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t v\u00e0 b\u1eadc hai\/ B\u00e0i 3: H\u00e0m s\u1ed1 b\u1eadc hai"},"content":{"rendered":"\n

Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 42<\/strong>: Nh\u1eafc l\u1ea1i c\u00e1c k\u1ebft qu\u1ea3 \u0111\u00e3 bi\u1ebft v\u1ec1 \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 y = ax2<\/sup>.<\/ins><\/p>\n\n\n\n

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

\u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = ax2<\/sup> l\u00e0 m\u1ed9t parabol:<\/p>\n\n\n\n

+ N\u1eb1m ph\u00eda tr\u00ean tr\u1ee5c ho\u00e0nh n\u1ebfu a > 0 v\u00e0 nh\u1eadn \u0111i\u1ec3m O(0;0) l\u00e0m \u0111i\u1ec3m th\u1ea5p nh\u1ea5t.<\/p>\n\n\n\n

+ N\u1eb1m ph\u00eda d\u01b0\u1edbi tr\u1ee5c ho\u00e0nh n\u1ebfu a < 0 v\u00e0 nh\u1eadn \u0111i\u1ec3m O(0;0) l\u00e0m \u0111i\u1ec3m cao nh\u1ea5t.<\/p>\n\n\n\n

Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 45<\/strong>: V\u1ebd parabol y = -2x^2 + x + 3.<\/ins><\/p>\n\n\n\n

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

\u0110\u1ec9nh I(1\/4; 25\/8)<\/p>\n\n\n\n

Tr\u1ee5c \u0111\u1ed1i x\u1ee9ng l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng x = 1\/4<\/p>\n\n\n\n

Giao \u0111i\u1ec3m v\u1edbi tr\u1ee5c Oy l\u00e0 \u0111i\u1ec3m (0;3)<\/p>\n\n\n\n

Giao \u0111i\u1ec3m v\u1edbi tr\u1ee5c Ox l\u00e0 \u0111i\u1ec3m (3\/2;0) v\u00e0 (-1;0)<\/p>\n\n\n\n

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

B\u00e0i 1 (trang 49 SGK \u0110\u1ea1i s\u1ed1 10):<\/strong> X\u00e1c \u0111\u1ecbnh t\u1ecda \u0111\u1ed9 c\u1ee7a \u0111\u1ec9nh v\u00e0 c\u00e1c giao \u0111i\u1ec3m v\u1edbi tr\u1ee5c tung, tr\u1ee5c ho\u00e0nh (n\u1ebfu c\u00f3) c\u1ee7a m\u1ed9t parabol:<\/p>\n\n\n\n

a) y = x2<\/sup> – 3x + 2 ;         b) y = -2x2<\/sup> + 4x – 3;<\/p>\n\n\n\n

c) y = x2<\/sup> – 2x ;             d) y = -x2<\/sup> + 4.<\/p>\n\n\n\n

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

a) y = x2<\/sup> \u2013 3x + 2 c\u00f3 a = 1 ; b = \t \u20133 ; c = 2 ; \u0394 = b2<\/sup> \u2013 4ac = (\u20133)2<\/sup> \u2013 4.2.1 = 1. <\/p>\n\n\n\n

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

+ \u0110\u1ec9nh c\u1ee7a Parabol l\u00e0 \n<\/p>\n\n\n\n

+ Khi x = 0 th\u00ec y = 2. V\u1eady giao \u0111i\u1ec3m v\u1edbi tr\u1ee5c tung l\u00e0 A(0 ; 2). <\/p>\n\n\n\n

+ Khi y = 0 th\u00ec x2<\/sup> \u2013 3x + 2 = 0. Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m x = 2 ho\u1eb7c x = 1. <\/p>\n\n\n\n

V\u1eady giao \u0111i\u1ec3m v\u1edbi tr\u1ee5c ho\u00e0nh l\u00e0 B(2 ; 0) v\u00e0 C(1 ; 0). <\/p>\n\n\n\n

b) y = \u20132x2<\/sup> + 4x \u2013 3 c\u00f3 a = \u20132 ; b = 4 ; c = \u20133 ; \u0394= b2<\/sup> \u2013 4ac = 42 \u2013 4.( \u20133).( \u20132) = \u20138<\/p>\n\n\n\n

+ \u0110\u1ec9nh c\u1ee7a Parabol l\u00e0 (1 ; \u20131). <\/ins><\/p>\n\n\n\n

+ Khi x = 0 th\u00ec y = \u20133. V\u1eady giao \u0111i\u1ec3m v\u1edbi tr\u1ee5c tung l\u00e0 A(0 ; \u20133). <\/p>\n\n\n\n

+ Khi y = 0 th\u00ec \u20132x2<\/sup> + 4x \u2013 3 = 0. Ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m. <\/p>\n\n\n\n

V\u1eady Parabol kh\u00f4ng c\u1eaft tr\u1ee5c ho\u00e0nh. <\/p>\n\n\n\n

c) y = x2<\/sup> \u2013 2x c\u00f3 a = 1 ; b = \u20132 ; c = 0 ; \u0394= b2<\/sup> \u2013 4ac = 4. <\/p>\n\n\n\n

+ \u0110\u1ec9nh c\u1ee7a Parabol l\u00e0 (1 ; \u20131). <\/p>\n\n\n\n

+ Khi x = 0 th\u00ec y = 0. V\u1eady giao \u0111i\u1ec3m v\u1edbi tr\u1ee5c tung l\u00e0 O(0 ; 0). <\/p>\n\n\n\n

+ Khi y = 0 th\u00ec x2<\/sup> \u2013 2x = 0. Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m x = 0 ho\u1eb7c x = 2. <\/p>\n\n\n\n

V\u1eady Parabol c\u1eaft tr\u1ee5c ho\u00e0nh t\u1ea1i hai \u0111i\u1ec3m O(0 ; 0) v\u00e0 A(2 ; 0). <\/ins><\/p>\n\n\n\n

d) y = \u2013x2<\/sup> + 4 c\u00f3 a = \u20131 ; b = 0 ; c = 4 ; \u0394= b2<\/sup> \u2013 4ac = 0 \u2013 4.( \u20131).4 = 16. <\/p>\n\n\n\n

+ \u0110\u1ec9nh c\u1ee7a Parabol l\u00e0 (0 ; 4). <\/p>\n\n\n\n

+ Khi x = 0 th\u00ec y = 4. V\u1eady giao \u0111i\u1ec3m v\u1edbi tr\u1ee5c tung l\u00e0 A(0 ; 4). <\/p>\n\n\n\n

+ Khi y = 0 th\u00ec \u2013x2<\/sup> + 4 = 0. Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m x = 2 ho\u1eb7c x = \u20132. <\/p>\n\n\n\n

V\u1eady Parabol c\u1eaft tr\u1ee5c ho\u00e0nh t\u1ea1i hai \u0111i\u1ec3m B(2 ; 0) ho\u1eb7c C(\u20132 ;0). <\/p>\n\n\n\n

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

+ Parabol y = ax2<\/sup> + bx + c c\u00f3 \u0111\u1ec9nh l\u00e0 I(\u2013b\/2a ; \u2013\u0394\/4a). <\/p>\n\n\n\n

B\u00e0i 2 (trang 49 SGK \u0110\u1ea1i s\u1ed1 10):<\/strong> L\u1eadp b\u1ea3ng bi\u1ebfn thi\u00ean v\u00e0 v\u1ebd \u0111\u1ed3 th\u1ecb c\u1ee7a c\u00e1c h\u00e0m s\u1ed1:<\/p>\n\n\n\n

a) y = 3x2<\/sup> – 4x + 1 ;         b) y = -3x2<\/sup> + 2x – 1<\/p>\n\n\n\n

c) y = 4x2<\/sup> – 4x + 1 ;         d) y = -x2<\/sup> + 4x – 4<\/ins><\/p>\n\n\n\n

e) y = 2x2<\/sup> + x + 1 ;          f) y = -x2<\/sup> + x – 1<\/p>\n\n\n\n

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

a) y = 3x2<\/sup> \u2013 4x + 1. <\/p>\n\n\n\n

+ T\u1eadp x\u00e1c \u0111\u1ecbnh: R. <\/p>\n\n\n\n

+ \u0110\u1ec9nh A(2\/3 ; \u20131\/3). <\/p>\n\n\n\n

+ Tr\u1ee5c \u0111\u1ed1i x\u1ee9ng x = 2\/3. <\/p>\n\n\n\n

+ Giao \u0111i\u1ec3m v\u1edbi Ox t\u1ea1i B(1\/3 ; 0) v\u00e0 C(1 ; 0). <\/p>\n\n\n\n

+ Giao \u0111i\u1ec3m v\u1edbi Oy t\u1ea1i D(0 ; 1). <\/p>\n\n\n\n

+ B\u1ea3ng bi\u1ebfn thi\u00ean:<\/p>\n\n\n\n

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

+ \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 :<\/p>\n\n\n\n

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

b) y = \u20133x2<\/sup> + 2x \u2013 1. <\/p>\n\n\n\n

+ T\u1eadp x\u00e1c \u0111\u1ecbnh: R<\/p>\n\n\n\n

+ \u0110\u1ec9nh A(1\/3 ; \u20132\/3). <\/p>\n\n\n\n

+ Tr\u1ee5c \u0111\u1ed1i x\u1ee9ng x = 1\/3. <\/p>\n\n\n\n

+ \u0110\u1ed3 th\u1ecb kh\u00f4ng giao v\u1edbi tr\u1ee5c ho\u00e0nh.<\/p>\n\n\n\n

+ Giao \u0111i\u1ec3m v\u1edbi tr\u1ee5c tung l\u00e0 B(0; \u20131). <\/p>\n\n\n\n

\u0110i\u1ec3m \u0111\u1ed1i x\u1ee9ng v\u1edbi B(0 ; \u20131) qua \u0111\u01b0\u1eddng th\u1eb3ng x = 1\/3 l\u00e0 C(2\/3 ; \u20131). <\/p>\n\n\n\n

+ B\u1ea3ng bi\u1ebfn thi\u00ean:<\/p>\n\n\n\n

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

+ \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 : <\/p>\n\n\n\n

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

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

c) y = 4x2<\/sup> \u2013 4x + 1. <\/p>\n\n\n\n

+ T\u1eadp x\u00e1c \u0111\u1ecbnh : R<\/p>\n\n\n\n

+ \u0110\u1ec9nh A(1\/2; 0). <\/p>\n\n\n\n

+ Tr\u1ee5c \u0111\u1ed1i x\u1ee9ng x = 1\/2. <\/p>\n\n\n\n

+ Giao \u0111i\u1ec3m v\u1edbi tr\u1ee5c ho\u00e0nh t\u1ea1i \u0111\u1ec9nh A.<\/p>\n\n\n\n

+ Giao \u0111i\u1ec3m v\u1edbi tr\u1ee5c tung B(0; 1).<\/p>\n\n\n\n

\u0110i\u1ec3m \u0111\u1ed1i x\u1ee9ng v\u1edbi B(0;1) qua \u0111\u01b0\u1eddng th\u1eb3ng x = 1\/2 l\u00e0 C(1; 1). <\/p>\n\n\n\n

+ B\u1ea3ng bi\u1ebfn thi\u00ean:<\/p>\n\n\n\n

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

+ \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1: <\/p>\n\n\n\n

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

d) y = \u2013x2<\/sup> + 4x \u2013 4. <\/p>\n\n\n\n

+ T\u1eadp x\u00e1c \u0111\u1ecbnh: R<\/p>\n\n\n\n

+ \u0110\u1ec9nh: I (2; 0)<\/p>\n\n\n\n

+ Tr\u1ee5c \u0111\u1ed1i x\u1ee9ng: x = 2.<\/p>\n\n\n\n

+ Giao \u0111i\u1ec3m v\u1edbi tr\u1ee5c ho\u00e0nh: A(2; 0).<\/p>\n\n\n\n

+ Giao \u0111i\u1ec3m v\u1edbi tr\u1ee5c tung: B(0; \u20134).<\/p>\n\n\n\n

\u0110i\u1ec3m \u0111\u1ed1i x\u1ee9ng v\u1edbi \u0111i\u1ec3m B(0; \u20134) qua \u0111\u01b0\u1eddng th\u1eb3ng x = 2 l\u00e0 C(4; \u20134). <\/p>\n\n\n\n

+ B\u1ea3ng bi\u1ebfn thi\u00ean:<\/p>\n\n\n\n

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

+ \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1: <\/p>\n\n\n\n

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

e) y = 2x2<\/sup> + x + 1<\/p>\n\n\n\n

+ T\u1eadp x\u00e1c \u0111\u1ecbnh: R<\/p>\n\n\n\n

+ \u0110\u1ec9nh A(\u20131\/4 ; 7\/8). <\/p>\n\n\n\n

+ Tr\u1ee5c \u0111\u1ed1i x\u1ee9ng x = \u20131\/4. <\/p>\n\n\n\n

+ \u0110\u1ed3 th\u1ecb kh\u00f4ng giao v\u1edbi tr\u1ee5c ho\u00e0nh.<\/p>\n\n\n\n

+ Giao \u0111i\u1ec3m v\u1edbi tr\u1ee5c tung B(0; 1).<\/p>\n\n\n\n

\u0110i\u1ec3m \u0111\u1ed1i x\u1ee9ng v\u1edbi B(0 ; 1) qua \u0111\u01b0\u1eddng th\u1eb3ng x = \u20131\/4 l\u00e0 C(\u20131\/2 ; 1)<\/p>\n\n\n\n

+ B\u1ea3ng bi\u1ebfn thi\u00ean:<\/p>\n\n\n\n

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

+ \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1: <\/p>\n\n\n\n

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

f) y = \u2013x2<\/sup> + x \u2013 1<\/p>\n\n\n\n

+ T\u1eadp x\u00e1c \u0111\u1ecbnh R<\/p>\n\n\n\n

+ \u0110\u1ec9nh A(1\/2 ; \u20133\/4). <\/p>\n\n\n\n

+ Tr\u1ee5c \u0111\u1ed1i x\u1ee9ng x = 1\/2. <\/p>\n\n\n\n

+ \u0110\u1ed3 th\u1ecb kh\u00f4ng giao v\u1edbi tr\u1ee5c ho\u00e0nh.<\/p>\n\n\n\n

+ Giao \u0111i\u1ec3m v\u1edbi tr\u1ee5c tung: B(0; \u20131).<\/p>\n\n\n\n

\u0110i\u1ec3m \u0111\u1ed1i x\u1ee9ng v\u1edbi B(0 ; \u20131) qua \u0111\u01b0\u1eddng th\u1eb3ng x = 1\/2 l\u00e0 C(1 ; \u20131). <\/p>\n\n\n\n

+ B\u1ea3ng bi\u1ebfn thi\u00ean:<\/p>\n\n\n\n

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

+ \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 : <\/p>\n\n\n\n

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

B\u00e0i 3 (trang 49 SGK \u0110\u1ea1i s\u1ed1 10):<\/strong> X\u00e1c \u0111\u1ecbnh parabol y = ax2<\/sup> + bx + 2, bi\u1ebft r\u1eb1ng parabol \u0111\u00f3:<\/p>\n\n\n\n

a) \u0110i qua hai \u0111i\u1ec3m M(1; 5) v\u00e0 N(-2; 8);<\/p>\n\n\n\n

b) \u0110i qua hai \u0111i\u1ec3m A(3; -4) v\u00e0 c\u00f3 tr\u1ee5c \u0111\u1ed1i x\u1ee9ng l\u00e0 x = -3\/2;<\/ins><\/p>\n\n\n\n

c) C\u00f3 \u0111\u1ec9nh l\u00e0 I(2; -2);<\/p>\n\n\n\n

d) \u0110i qua \u0111i\u1ec3m B(-1; 6) v\u00e0 tung \u0111\u1ed9 c\u1ee7a \u0111\u1ec9nh l\u00e0 -1\/4.<\/p>\n\n\n\n

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

a) <\/p>\n\n\n\n

+ Parabol y = ax2<\/sup> + bx + 2 \u0111i qua M(1 ; 5) <\/p>\n\n\n\n

\u21d2 5 = a.12 + b.1 + 2 \u21d2 a + b = 3 (1) . <\/p>\n\n\n\n

+ Parabol y = ax2<\/sup> + bx + 2 \u0111i qua N(\u20132; 8)<\/p>\n\n\n\n

\u21d2 8 = a.( \u20132)2<\/sup> + b.( \u20132) + 2 \u21d2 4a \u2013 2b = 6 (2). <\/p>\n\n\n\n

T\u1eeb (1) v\u00e0 (2) suy ra: a = 2; b = 1. <\/p>\n\n\n\n

V\u1eady parabol c\u1ea7n t\u00ecm l\u00e0 y = 2x2<\/sup> + x + 2. <\/p>\n\n\n\n

b) + Parabol y = ax2<\/sup> + bx + 2 c\u00f3 tr\u1ee5c \u0111\u1ed1i x\u1ee9ng x = \u20133\/2 <\/p>\n\n\n\n

\u21d2 \u2013b\/2a = \u20133\/2 \u21d2 b = 3a (1)<\/p>\n\n\n\n

+ Parabol y = ax2<\/sup> + bx + 2 \u0111i qua \u0111i\u1ec3m A(3; \u20134) <\/p>\n\n\n\n

\u21d2 \u20134 = a.32<\/sup> + b.3 + 2 \u21d2 9a + 3b = \u20136 (2). <\/p>\n\n\n\n

Thay b = 3a \u1edf (1) v\u00e0o bi\u1ec3u th\u1ee9c (2) ta \u0111\u01b0\u1ee3c: <\/p>\n\n\n\n

9a + 3.3a = \u20136 \u21d2 18a = \u20136 \u21d2 a = \u20131\/3 \u21d2 b = \u20131. <\/p>\n\n\n\n

V\u1eady parabol c\u1ea7n t\u00ecm l\u00e0 y = \u20131\/3x2<\/sup> \u2013 x + 2. <\/ins><\/p>\n\n\n\n

c) Parabol y = ax2<\/sup> + bx + 2 c\u00f3 \u0111\u1ec9nh I(2 ; \u20132), suy ra : <\/p>\n\n\n\n

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

T\u1eeb (1) \u21d2 b2<\/sup> = 16.a2<\/sup>, thay v\u00e0o (2) ta \u0111\u01b0\u1ee3c 16a2<\/sup> = 16a \u21d2 a = 1 \u21d2 b = \u20134. <\/p>\n\n\n\n

V\u1eady parabol c\u1ea7n t\u00ecm l\u00e0 y = x2<\/sup> \u2013 4x + 2. <\/p>\n\n\n\n

d) + Parabol y = ax2<\/sup> + bx + 2 \u0111i qua \u0111i\u1ec3m B(\u20131 ; 6) <\/p>\n\n\n\n

\u21d2 6 = a.( \u20131)2<\/sup> + b.( \u20131) + 2 \u21d2 a = b + 4 (1) <\/p>\n\n\n\n

+ Parabol y = ax2<\/sup> + bx + 2 c\u00f3 tung \u0111\u1ed9 c\u1ee7a \u0111\u1ec9nh l\u00e0 \u20131\/4<\/p>\n\n\n\n

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

Thay (1) v\u00e0o (2) ta \u0111\u01b0\u1ee3c: b2<\/sup> = 9.(b + 4) \u21d4 b2<\/sup> \u2013 9b \u2013 36 = 0. <\/p>\n\n\n\n

Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m b = 12 ho\u1eb7c b = \u20133. <\/p>\n\n\n\n

V\u1edbi b = 12 th\u00ec a = 16. <\/p>\n\n\n\n

V\u1edbi b = \u20133 th\u00ec a = 1. <\/p>\n\n\n\n

V\u1eady c\u00f3 hai parabol th\u1ecfa m\u00e3n l\u00e0 y = 16x2<\/sup> + 12b + 2 v\u00e0 y = x2<\/sup> \u2013 3x + 2. <\/p>\n\n\n\n

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

Parabol y = ax2<\/sup> + bx + c c\u00f3 : <\/p>\n\n\n\n

+ \u0110\u1ec9nh l\u00e0 I(\u2013b\/2a ; \u2013 \u0394\/4a)<\/p>\n\n\n\n

+ Tr\u1ee5c \u0111\u1ed1i x\u1ee9ng l\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng x = \u2013b\/2a <\/p>\n\n\n\n

B\u00e0i 4 (trang 50 SGK \u0110\u1ea1i s\u1ed1 10):<\/strong> X\u00e1c \u0111\u1ecbnh a, b, c bi\u1ebft parabol y = ax2<\/sup> + bx + c \u0111i qua \u0111i\u1ec3m A(8 ; 0) v\u00e0 c\u00f3 \u0111\u1ec9nh l\u00e0 I(6 ; -12).<\/p>\n\n\n\n

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

+ Parabol y = ax2<\/sup> + bx + c \u0111i qua \u0111i\u1ec3m A (8; 0) <\/p>\n\n\n\n

\u21d2 0 = a.82<\/sup> + b.8 + c \u21d2 64a + 8b + c = 0 (1). <\/p>\n\n\n\n

+ Parabol y = ax2<\/sup> + bx + c c\u00f3 \u0111\u1ec9nh l\u00e0 I (6 ; \u201312) suy ra: <\/p>\n\n\n\n

\u2013b\/2a = 6 \u21d2 b = \u201312a (2). <\/p>\n\n\n\n

\u2013\u0394\/4a = \u201312 \u21d2 \u0394 = 48a \u21d2 b2<\/sup> \u2013 4ac = 48a (3) . <\/p>\n\n\n\n

Thay (2) v\u00e0o (1) ta c\u00f3: 64a \u2013 96a + c = 0 \u21d2 c = 32a. <\/p>\n\n\n\n

Thay b = \u201312a v\u00e0 c = 32a v\u00e0o (3) ta \u0111\u01b0\u1ee3c: <\/p>\n\n\n\n

(\u201312a)2<\/sup> \u2013 4a.32a = 48a <\/p>\n\n\n\n

\u21d2 144a2<\/sup> \u2013 128a2<\/sup> = 48a <\/p>\n\n\n\n

\u21d2 16a2<\/sup> = 48a <\/p>\n\n\n\n

\u21d2 a = 3 (v\u00ec a \u2260 0). <\/p>\n\n\n\n

T\u1eeb a = 3 \u21d2 b = \u201336 v\u00e0 c = 96. <\/p>\n\n\n\n

V\u1eady a = 3; b = \u201336 v\u00e0 c = 96.<\/p>\n","protected":false},"excerpt":{"rendered":"

Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 42: Nh\u1eafc l\u1ea1i c\u00e1c k\u1ebft qu\u1ea3 \u0111\u00e3 bi\u1ebft v\u1ec1 \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 y = ax2. L\u1eddi gi\u1ea3i \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = ax2 l\u00e0 m\u1ed9t parabol: + N\u1eb1m ph\u00eda tr\u00ean tr\u1ee5c ho\u00e0nh n\u1ebfu a > 0 v\u00e0 nh\u1eadn \u0111i\u1ec3m O(0;0) l\u00e0m […]<\/p>\n","protected":false},"author":12,"featured_media":44321,"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 2: H\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t v\u00e0 b\u1eadc hai\/ B\u00e0i 3: H\u00e0m s\u1ed1 b\u1eadc hai<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/lop12.edu.vn\/giai-toan-10-chuong-2-ham-so-bac-nhat-va-bac-hai-bai-3-ham-so-bac-hai\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 2: H\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t v\u00e0 b\u1eadc hai\/ B\u00e0i 3: H\u00e0m s\u1ed1 b\u1eadc hai\" \/>\n<meta property=\"og:description\" content=\"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 42: Nh\u1eafc l\u1ea1i c\u00e1c k\u1ebft qu\u1ea3 \u0111\u00e3 bi\u1ebft v\u1ec1 \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 y = ax2. L\u1eddi gi\u1ea3i \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = ax2 l\u00e0 m\u1ed9t parabol: + N\u1eb1m ph\u00eda tr\u00ean tr\u1ee5c ho\u00e0nh n\u1ebfu a > 0 v\u00e0 nh\u1eadn \u0111i\u1ec3m O(0;0) l\u00e0m […]\" \/>\n<meta property=\"og:url\" content=\"https:\/\/lop12.edu.vn\/giai-toan-10-chuong-2-ham-so-bac-nhat-va-bac-hai-bai-3-ham-so-bac-hai\/\" \/>\n<meta property=\"og:site_name\" content=\"Lop12.edu.vn - C\u1ed9ng \u0111\u1ed3ng h\u1ecdc sinh l\u1edbp 12 l\u1edbn nh\u1ea5t t\u1ea1i Vi\u1ec7t Nam\" \/>\n<meta property=\"article:published_time\" content=\"2019-10-15T07:14:16+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2019-10-15T07:14:18+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/lop12.edu.vn\/wp-content\/uploads\/2019\/10\/h4.png\" \/>\n\t<meta property=\"og:image:width\" content=\"222\" \/>\n\t<meta property=\"og:image:height\" content=\"302\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/png\" \/>\n<meta name=\"author\" content=\"Nguy\u1ec5n M\u01a1\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Nguy\u1ec5n M\u01a1\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"7 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"WebPage\",\"@id\":\"https:\/\/lop12.edu.vn\/giai-toan-10-chuong-2-ham-so-bac-nhat-va-bac-hai-bai-3-ham-so-bac-hai\/\",\"url\":\"https:\/\/lop12.edu.vn\/giai-toan-10-chuong-2-ham-so-bac-nhat-va-bac-hai-bai-3-ham-so-bac-hai\/\",\"name\":\"[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 2: H\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t v\u00e0 b\u1eadc hai\/ B\u00e0i 3: H\u00e0m s\u1ed1 b\u1eadc hai\",\"isPartOf\":{\"@id\":\"https:\/\/lop12.edu.vn\/#website\"},\"datePublished\":\"2019-10-15T07:14:16+00:00\",\"dateModified\":\"2019-10-15T07:14:18+00:00\",\"author\":{\"@id\":\"https:\/\/lop12.edu.vn\/#\/schema\/person\/104e47bfb6189ee6fa1aa67e1f9107a2\"},\"breadcrumb\":{\"@id\":\"https:\/\/lop12.edu.vn\/giai-toan-10-chuong-2-ham-so-bac-nhat-va-bac-hai-bai-3-ham-so-bac-hai\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/lop12.edu.vn\/giai-toan-10-chuong-2-ham-so-bac-nhat-va-bac-hai-bai-3-ham-so-bac-hai\/\"]}]},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/lop12.edu.vn\/giai-toan-10-chuong-2-ham-so-bac-nhat-va-bac-hai-bai-3-ham-so-bac-hai\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/lop12.edu.vn\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 2: H\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t v\u00e0 b\u1eadc hai\/ B\u00e0i 3: H\u00e0m s\u1ed1 b\u1eadc hai\"}]},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/lop12.edu.vn\/#website\",\"url\":\"https:\/\/lop12.edu.vn\/\",\"name\":\"Lop12.edu.vn - C\u1ed9ng \u0111\u1ed3ng h\u1ecdc sinh l\u1edbp 12\",\"description\":\"\",\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/lop12.edu.vn\/?s={search_term_string}\"},\"query-input\":\"required name=search_term_string\"}],\"inLanguage\":\"en-US\"},{\"@type\":\"Person\",\"@id\":\"https:\/\/lop12.edu.vn\/#\/schema\/person\/104e47bfb6189ee6fa1aa67e1f9107a2\",\"name\":\"Nguy\u1ec5n M\u01a1\",\"image\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/lop12.edu.vn\/#\/schema\/person\/image\/\",\"url\":\"https:\/\/secure.gravatar.com\/avatar\/20d6905502209505aa7a21b55419ebe9?s=96&d=mm&r=g\",\"contentUrl\":\"https:\/\/secure.gravatar.com\/avatar\/20d6905502209505aa7a21b55419ebe9?s=96&d=mm&r=g\",\"caption\":\"Nguy\u1ec5n M\u01a1\"},\"url\":\"https:\/\/lop12.edu.vn\/author\/mont\/\"}]}<\/script>\n<!-- \/ Yoast SEO plugin. -->","yoast_head_json":{"title":"[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 2: H\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t v\u00e0 b\u1eadc hai\/ B\u00e0i 3: H\u00e0m s\u1ed1 b\u1eadc hai","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/lop12.edu.vn\/giai-toan-10-chuong-2-ham-so-bac-nhat-va-bac-hai-bai-3-ham-so-bac-hai\/","og_locale":"en_US","og_type":"article","og_title":"[Gi\u1ea3i To\u00e1n 10] Ch\u01b0\u01a1ng 2: H\u00e0m s\u1ed1 b\u1eadc nh\u1ea5t v\u00e0 b\u1eadc hai\/ B\u00e0i 3: H\u00e0m s\u1ed1 b\u1eadc hai","og_description":"Tr\u1ea3 l\u1eddi c\u00e2u h\u1ecfi To\u00e1n 10 \u0110\u1ea1i s\u1ed1 B\u00e0i 3 trang 42: Nh\u1eafc l\u1ea1i c\u00e1c k\u1ebft qu\u1ea3 \u0111\u00e3 bi\u1ebft v\u1ec1 \u0111\u1ed3 th\u1ecb c\u1ee7a h\u00e0m s\u1ed1 y = ax2. 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